LeetCode #3505 — HARD

Minimum Operations to Make Elements Within K Subarrays Equal

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode

The Problem

Problem Statement

You are given an integer array nums and two integers, x and k. You can perform the following operation any number of times (including zero):

Increase or decrease any element of nums by 1.

Return the minimum number of operations needed to have at least k non-overlapping subarrays of size exactly x in nums, where all elements within each subarray are equal.

Example 1:

Input: nums = [5,-2,1,3,7,3,6,4,-1], x = 3, k = 2

Output: 8

Explanation:

Use 3 operations to add 3 to nums[1] and use 2 operations to subtract 2 from nums[3]. The resulting array is [5, 1, 1, 1, 7, 3, 6, 4, -1].
Use 1 operation to add 1 to nums[5] and use 2 operations to subtract 2 from nums[6]. The resulting array is [5, 1, 1, 1, 7, 4, 4, 4, -1].
Now, all elements within each subarray [1, 1, 1] (from indices 1 to 3) and [4, 4, 4] (from indices 5 to 7) are equal. Since 8 total operations were used, 8 is the output.

Example 2:

Input: nums = [9,-2,-2,-2,1,5], x = 2, k = 2

Output: 3

Explanation:

Use 3 operations to subtract 3 from nums[4]. The resulting array is [9, -2, -2, -2, -2, 5].
Now, all elements within each subarray [-2, -2] (from indices 1 to 2) and [-2, -2] (from indices 3 to 4) are equal. Since 3 operations were used, 3 is the output.

Constraints:

2 <= nums.length <= 10⁵
-10⁶ <= nums[i] <= 10⁶
2 <= x <= nums.length
1 <= k <= 15
2 <= k * x <= nums.length

Roadmap

Brute Force Baseline
Core Insight
Algorithm Walkthrough
Edge Cases
Full Annotated Code
Interactive Study Demo
Complexity Analysis

Step 01

Brute Force Baseline

Problem summary: You are given an integer array nums and two integers, x and k. You can perform the following operation any number of times (including zero): Increase or decrease any element of nums by 1. Return the minimum number of operations needed to have at least k non-overlapping subarrays of size exactly x in nums, where all elements within each subarray are equal.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map · Math · Dynamic Programming · Sliding Window

Example 1

[5,-2,1,3,7,3,6,4,-1]
3
2

Example 2

[9,-2,-2,-2,1,5]
2
2

Related Problems

Find Median from Data Stream (find-median-from-data-stream)
Minimum Moves to Equal Array Elements II (minimum-moves-to-equal-array-elements-ii)

Step 02

Core Insight

What unlocks the optimal approach

Making every element of an x-sized window equal to its median is optimal.
Precalculate this for each window.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3505: Minimum Operations to Make Elements Within K Subarrays Equal
class MyMap {
  public TreeMap<Integer, Integer> map = new TreeMap<>();
  public int size = 0;
  public long sum = 0;
}

class Solution {
  public long minOperations(int[] nums, int x, int k) {
    // minOps[i] := the minimum number of operations needed to make
    // nums[i..i + x - 1] equal to the median
    List<Long> minOps = getMinOps(nums, x);
    Long[][] mem = new Long[nums.length + 1][k + 1];
    return minOperations(nums, x, 0, k, minOps, mem);
  }

  private static final long INF = Long.MAX_VALUE / 2;

  // Returns the minimum operations needed to have at least k non-overlapping
  // subarrays of size x in nums[i..n - 1].
  private long minOperations(int[] nums, int x, int i, int k, List<Long> minOps, Long[][] mem) {
    if (k == 0)
      return 0;
    if (i == nums.length)
      return INF;
    if (mem[i][k] != null)
      return mem[i][k];
    final long skip = minOperations(nums, x, i + 1, k, minOps, mem);
    final long pick = i + x <= nums.length
                          ? minOps.get(i) + minOperations(nums, x, i + x, k - 1, minOps, mem)
                          : INF;
    return mem[i][k] = Math.min(skip, pick);
  }

  // Returns the minimum operations needed to make all elements in the window of
  // size x equal to the median.
  private List<Long> getMinOps(int[] nums, int x) {
    List<Long> minOps = new ArrayList<>();
    MyMap lower = new MyMap();
    MyMap upper = new MyMap();
    for (int i = 0; i < nums.length; ++i) {
      if (lower.map.isEmpty() || nums[i] <= lower.map.lastKey()) {
        lower.map.merge(nums[i], 1, Integer::sum);
        lower.sum += nums[i];
        ++lower.size;
      } else {
        upper.map.merge(nums[i], 1, Integer::sum);
        upper.sum += nums[i];
        ++upper.size;
      }
      if (i >= x) {
        final int outNum = nums[i - x];
        if (lower.map.containsKey(outNum)) {
          lower.map.merge(outNum, -1, Integer::sum);
          if (lower.map.get(outNum) == 0)
            lower.map.remove(outNum);
          lower.sum -= outNum;
          --lower.size;
        } else {
          upper.map.merge(outNum, -1, Integer::sum);
          if (upper.map.get(outNum) == 0)
            upper.map.remove(outNum);
          upper.sum -= outNum;
          --upper.size;
        }
      }
      // Balance the two maps s.t.
      // |lower| >= |upper| and |lower| - |upper| <= 1.
      if (lower.size < upper.size) {
        final int val = upper.map.firstKey();
        upper.map.merge(val, -1, Integer::sum);
        if (upper.map.get(val) == 0)
          upper.map.remove(val);
        lower.map.merge(val, 1, Integer::sum);
        upper.sum -= val;
        lower.sum += val;
        --upper.size;
        ++lower.size;
      } else if (lower.size - upper.size > 1) {
        final int val = lower.map.lastKey();
        lower.map.merge(val, -1, Integer::sum);
        if (lower.map.get(val) == 0)
          lower.map.remove(val);
        upper.map.merge(val, 1, Integer::sum);
        lower.sum -= val;
        upper.sum += val;
        --lower.size;
        ++upper.size;
      }
      // Calculate operations needed to make all elements in the window equal
      // to the median.
      if (i >= x - 1) {
        final int median = lower.map.lastKey();
        final long ops = (median * lower.size - lower.sum) + (upper.sum - median * upper.size);
        minOps.add(ops);
      }
    }
    return minOps;
  }
}

// Accepted solution for LeetCode #3505: Minimum Operations to Make Elements Within K Subarrays Equal
package main

import (
	"container/heap"
	"math"
	"sort"
)

// https://space.bilibili.com/206214
func minOperations1(nums []int, x, k int) int64 {
	n := len(nums)
	dis := medianSlidingWindow(nums, x)
	f := make([][]int, k+1)
	for i := range f {
		f[i] = make([]int, n+1)
	}
	for i := 1; i <= k; i++ {
		f[i][i*x-1] = math.MaxInt
		for j := i * x; j <= n-(k-i)*x; j++ { // 左右留出足够空间给其他子数组
			f[i][j] = min(f[i][j-1], f[i-1][j-x]+dis[j-x]) // j-x 为子数组左端点
		}
	}
	return int64(f[k][n])
}

func minOperations(nums []int, x, k int) int64 {
	n := len(nums)
	dis := medianSlidingWindow(nums, x)
	f := make([]int, n+1)
	g := make([]int, n+1) // 滚动数组
	for i := 1; i <= k; i++ {
		g[i*x-1] = math.MaxInt
		for j := i * x; j <= n-(k-i)*x; j++ { // 左右留出足够空间给其他子数组
			g[j] = min(g[j-1], f[j-x]+dis[j-x]) // j-x 为子数组左端点
		}
		f, g = g, f
	}
	return int64(f[n])
}

// 480. 滑动窗口中位数（有改动）
// 返回 nums 的所有长为 k 的子数组的（到子数组中位数的）距离和
func medianSlidingWindow(nums []int, k int) []int {
	ans := make([]int, len(nums)-k+1)
	left := newLazyHeap()  // 最大堆（元素取反）
	right := newLazyHeap() // 最小堆

	for i, in := range nums {
		// 1. 进入窗口
		if left.size == right.size {
			left.push(-right.pushPop(in))
		} else {
			right.push(-left.pushPop(-in))
		}

		l := i + 1 - k
		if l < 0 { // 窗口大小不足 k
			continue
		}

		// 2. 计算答案
		v := -left.top()
		s1 := v*left.size + left.sum // sum 取反
		s2 := right.sum - v*right.size
		ans[l] = s1 + s2

		// 3. 离开窗口
		out := nums[l]
		if out <= -left.top() {
			left.remove(-out)
			if left.size < right.size {
				left.push(-right.pop()) // 平衡两个堆的大小
			}
		} else {
			right.remove(out)
			if left.size > right.size+1 {
				right.push(-left.pop()) // 平衡两个堆的大小
			}
		}
	}

	return ans
}

func newLazyHeap() *lazyHeap {
	return &lazyHeap{removeCnt: map[int]int{}}
}

// 懒删除堆
type lazyHeap struct {
	sort.IntSlice
	removeCnt map[int]int // 每个元素剩余需要删除的次数
	size      int         // 实际大小
	sum       int         // 堆中元素总和
}

// 必须实现的两个接口
func (h *lazyHeap) Push(v any) { h.IntSlice = append(h.IntSlice, v.(int)) }
func (h *lazyHeap) Pop() any   { a := h.IntSlice; v := a[len(a)-1]; h.IntSlice = a[:len(a)-1]; return v }

// 删除
func (h *lazyHeap) remove(v int) {
	h.removeCnt[v]++ // 懒删除
	h.size--
	h.sum -= v
}

// 正式执行删除操作
func (h *lazyHeap) applyRemove() {
	for h.removeCnt[h.IntSlice[0]] > 0 {
		h.removeCnt[h.IntSlice[0]]--
		heap.Pop(h)
	}
}

// 查看堆顶
func (h *lazyHeap) top() int {
	h.applyRemove()
	return h.IntSlice[0]
}

// 出堆
func (h *lazyHeap) pop() int {
	h.applyRemove()
	h.size--
	h.sum -= h.IntSlice[0]
	return heap.Pop(h).(int)
}

// 入堆
func (h *lazyHeap) push(v int) {
	if h.removeCnt[v] > 0 {
		h.removeCnt[v]-- // 抵消之前的删除
	} else {
		heap.Push(h, v)
	}
	h.size++
	h.sum += v
}

// push(v) 然后 pop()
func (h *lazyHeap) pushPop(v int) int {
	if h.size > 0 && v > h.top() { // 最小堆，v 比堆顶大就替换堆顶
		h.sum += v - h.IntSlice[0]
		v, h.IntSlice[0] = h.IntSlice[0], v
		heap.Fix(h, 0)
	}
	return v
}

# Accepted solution for LeetCode #3505: Minimum Operations to Make Elements Within K Subarrays Equal
# Time:  O(nlogx + k * n)
# Space: O(n)

from sortedcontainers import SortedList


# two sorted lists, dp
class Solution(object):
    def minOperations(self, nums, x, k):
        """
        :type nums: List[int]
        :type x: int
        :type k: int
        :rtype: int
        """
        class SlidingWindow(object):
            def __init__(self):
                self.left = SortedList()
                self.right = SortedList()
                self.total1 = self.total2 = 0

            def add(self, val):
                if not self.left or val <= self.left[-1]:
                    self.left.add(val)
                    self.total1 += val
                else:
                    self.right.add(val)
                    self.total2 += val
                self.rebalance()

            def remove(self, val):
                if val <= self.left[-1]:
                    self.left.remove(val)
                    self.total1 -= val
                else:
                    self.right.remove(val)
                    self.total2 -= val
                self.rebalance()

            def rebalance(self):
                if len(self.left) < len(self.right):
                    self.total2 -= self.right[0]
                    self.total1 += self.right[0]
                    self.left.add(self.right[0])
                    self.right.pop(0)
                elif len(self.left) > len(self.right)+1:
                    self.total1 -= self.left[-1]
                    self.total2 += self.left[-1]
                    self.right.add(self.left[-1])
                    self.left.pop()

            def median(self):
                return self.left[-1]


        INF = float("inf")
        sw = SlidingWindow()
        cost = [INF]*(len(nums)+1)
        for i in xrange(len(nums)):
            if i-x >= 0:
                sw.remove(nums[i-x])
            sw.add(nums[i])
            if i >= x-1:
                cost[i+1] = (sw.median()*len(sw.left)-sw.total1) + (sw.total2-sw.median()*len(sw.right))
        dp = [0]*(len(nums)+1)
        for i in xrange(k):
            new_dp = [INF]*(len(nums)+1)
            for j in xrange((i+1)*x, len(nums)+1):
                new_dp[j] = min(new_dp[j-1], dp[j-x]+cost[j])
            dp = new_dp
        return dp[-1]


# Time:  O(nlogx + k * n)
# Space: O(n)
import heapq
import collections


# two heaps, dp
class Solution2(object):
    def minOperations(self, nums, x, k):
        """
        :type nums: List[int]
        :type x: int
        :type k: int
        :rtype: int
        """
        class LazyHeap(object):
            def __init__(self, sign):
                self.heap = []
                self.to_remove = collections.defaultdict(int)
                self.cnt = 0
                self.sign = sign

            def push(self, val):
                heapq.heappush(self.heap, self.sign*val)

            def full_remove(self):
                result = []
                for x in self.heap:
                    if x not in self.to_remove:
                        result.append(x)
                        continue
                    self.to_remove[x] -= 1
                    if not self.to_remove[x]:
                        del self.to_remove[x]
                self.heap[:] = result
                heapq.heapify(self.heap)
    
            def remove(self, val):
                self.to_remove[self.sign*val] += 1
                self.cnt += 1
                if self.cnt > len(self.heap)-self.cnt:
                    self.full_remove()
                    self.cnt = 0

            def pop(self):
                self.remove(self.top())

            def top(self):
                while self.heap and self.heap[0] in self.to_remove:
                    self.to_remove[self.heap[0]] -= 1
                    self.cnt -= 1
                    if self.to_remove[self.heap[0]] == 0:
                        del self.to_remove[self.heap[0]]
                    heapq.heappop(self.heap)
                return self.sign*self.heap[0]

            def __len__(self):
                return len(self.heap)-self.cnt


        class SlidingWindow(object):
            def __init__(self):
                self.left = LazyHeap(-1)   # max heap
                self.right = LazyHeap(+1)  # min heap
                self.total1 = self.total2 = 0

            def add(self, val):
                if not self.left or val <= self.left.top():
                    self.left.push(val)
                    self.total1 += val
                else:
                    self.right.push(val)
                    self.total2 += val
                self.rebalance()

            def remove(self, val):
                if val <= self.left.top():
                    self.left.remove(val)
                    self.total1 -= val
                else:
                    self.right.remove(val)
                    self.total2 -= val
                self.rebalance()

            def rebalance(self):
                if len(self.left) < len(self.right):
                    self.total2 -= self.right.top()
                    self.total1 += self.right.top()
                    self.left.push(self.right.top())
                    self.right.pop()
                elif len(self.left) > len(self.right)+1:
                    self.total1 -= self.left.top()
                    self.total2 += self.left.top()
                    self.right.push(self.left.top())
                    self.left.pop()

            def median(self):
                return self.left.top()


        INF = float("inf")
        sw = SlidingWindow()
        cost = [INF]*(len(nums)+1)
        for i in xrange(len(nums)):
            if i-x >= 0:
                sw.remove(nums[i-x])
            sw.add(nums[i])
            if i >= x-1:
                cost[i+1] = (sw.median()*len(sw.left)-sw.total1) + (sw.total2-sw.median()*len(sw.right))
        dp = [0]*(len(nums)+1)
        for i in xrange(k):
            new_dp = [INF]*(len(nums)+1)
            for j in xrange((i+1)*x, len(nums)+1):
                new_dp[j] = min(new_dp[j-1], dp[j-x]+cost[j])
            dp = new_dp
        return dp[-1]

// Accepted solution for LeetCode #3505: Minimum Operations to Make Elements Within K Subarrays Equal
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3505: Minimum Operations to Make Elements Within K Subarrays Equal
// class MyMap {
//   public TreeMap<Integer, Integer> map = new TreeMap<>();
//   public int size = 0;
//   public long sum = 0;
// }
// 
// class Solution {
//   public long minOperations(int[] nums, int x, int k) {
//     // minOps[i] := the minimum number of operations needed to make
//     // nums[i..i + x - 1] equal to the median
//     List<Long> minOps = getMinOps(nums, x);
//     Long[][] mem = new Long[nums.length + 1][k + 1];
//     return minOperations(nums, x, 0, k, minOps, mem);
//   }
// 
//   private static final long INF = Long.MAX_VALUE / 2;
// 
//   // Returns the minimum operations needed to have at least k non-overlapping
//   // subarrays of size x in nums[i..n - 1].
//   private long minOperations(int[] nums, int x, int i, int k, List<Long> minOps, Long[][] mem) {
//     if (k == 0)
//       return 0;
//     if (i == nums.length)
//       return INF;
//     if (mem[i][k] != null)
//       return mem[i][k];
//     final long skip = minOperations(nums, x, i + 1, k, minOps, mem);
//     final long pick = i + x <= nums.length
//                           ? minOps.get(i) + minOperations(nums, x, i + x, k - 1, minOps, mem)
//                           : INF;
//     return mem[i][k] = Math.min(skip, pick);
//   }
// 
//   // Returns the minimum operations needed to make all elements in the window of
//   // size x equal to the median.
//   private List<Long> getMinOps(int[] nums, int x) {
//     List<Long> minOps = new ArrayList<>();
//     MyMap lower = new MyMap();
//     MyMap upper = new MyMap();
//     for (int i = 0; i < nums.length; ++i) {
//       if (lower.map.isEmpty() || nums[i] <= lower.map.lastKey()) {
//         lower.map.merge(nums[i], 1, Integer::sum);
//         lower.sum += nums[i];
//         ++lower.size;
//       } else {
//         upper.map.merge(nums[i], 1, Integer::sum);
//         upper.sum += nums[i];
//         ++upper.size;
//       }
//       if (i >= x) {
//         final int outNum = nums[i - x];
//         if (lower.map.containsKey(outNum)) {
//           lower.map.merge(outNum, -1, Integer::sum);
//           if (lower.map.get(outNum) == 0)
//             lower.map.remove(outNum);
//           lower.sum -= outNum;
//           --lower.size;
//         } else {
//           upper.map.merge(outNum, -1, Integer::sum);
//           if (upper.map.get(outNum) == 0)
//             upper.map.remove(outNum);
//           upper.sum -= outNum;
//           --upper.size;
//         }
//       }
//       // Balance the two maps s.t.
//       // |lower| >= |upper| and |lower| - |upper| <= 1.
//       if (lower.size < upper.size) {
//         final int val = upper.map.firstKey();
//         upper.map.merge(val, -1, Integer::sum);
//         if (upper.map.get(val) == 0)
//           upper.map.remove(val);
//         lower.map.merge(val, 1, Integer::sum);
//         upper.sum -= val;
//         lower.sum += val;
//         --upper.size;
//         ++lower.size;
//       } else if (lower.size - upper.size > 1) {
//         final int val = lower.map.lastKey();
//         lower.map.merge(val, -1, Integer::sum);
//         if (lower.map.get(val) == 0)
//           lower.map.remove(val);
//         upper.map.merge(val, 1, Integer::sum);
//         lower.sum -= val;
//         upper.sum += val;
//         --lower.size;
//         ++upper.size;
//       }
//       // Calculate operations needed to make all elements in the window equal
//       // to the median.
//       if (i >= x - 1) {
//         final int median = lower.map.lastKey();
//         final long ops = (median * lower.size - lower.sum) + (upper.sum - median * upper.size);
//         minOps.add(ops);
//       }
//     }
//     return minOps;
//   }
// }

// Accepted solution for LeetCode #3505: Minimum Operations to Make Elements Within K Subarrays Equal
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #3505: Minimum Operations to Make Elements Within K Subarrays Equal
// class MyMap {
//   public TreeMap<Integer, Integer> map = new TreeMap<>();
//   public int size = 0;
//   public long sum = 0;
// }
// 
// class Solution {
//   public long minOperations(int[] nums, int x, int k) {
//     // minOps[i] := the minimum number of operations needed to make
//     // nums[i..i + x - 1] equal to the median
//     List<Long> minOps = getMinOps(nums, x);
//     Long[][] mem = new Long[nums.length + 1][k + 1];
//     return minOperations(nums, x, 0, k, minOps, mem);
//   }
// 
//   private static final long INF = Long.MAX_VALUE / 2;
// 
//   // Returns the minimum operations needed to have at least k non-overlapping
//   // subarrays of size x in nums[i..n - 1].
//   private long minOperations(int[] nums, int x, int i, int k, List<Long> minOps, Long[][] mem) {
//     if (k == 0)
//       return 0;
//     if (i == nums.length)
//       return INF;
//     if (mem[i][k] != null)
//       return mem[i][k];
//     final long skip = minOperations(nums, x, i + 1, k, minOps, mem);
//     final long pick = i + x <= nums.length
//                           ? minOps.get(i) + minOperations(nums, x, i + x, k - 1, minOps, mem)
//                           : INF;
//     return mem[i][k] = Math.min(skip, pick);
//   }
// 
//   // Returns the minimum operations needed to make all elements in the window of
//   // size x equal to the median.
//   private List<Long> getMinOps(int[] nums, int x) {
//     List<Long> minOps = new ArrayList<>();
//     MyMap lower = new MyMap();
//     MyMap upper = new MyMap();
//     for (int i = 0; i < nums.length; ++i) {
//       if (lower.map.isEmpty() || nums[i] <= lower.map.lastKey()) {
//         lower.map.merge(nums[i], 1, Integer::sum);
//         lower.sum += nums[i];
//         ++lower.size;
//       } else {
//         upper.map.merge(nums[i], 1, Integer::sum);
//         upper.sum += nums[i];
//         ++upper.size;
//       }
//       if (i >= x) {
//         final int outNum = nums[i - x];
//         if (lower.map.containsKey(outNum)) {
//           lower.map.merge(outNum, -1, Integer::sum);
//           if (lower.map.get(outNum) == 0)
//             lower.map.remove(outNum);
//           lower.sum -= outNum;
//           --lower.size;
//         } else {
//           upper.map.merge(outNum, -1, Integer::sum);
//           if (upper.map.get(outNum) == 0)
//             upper.map.remove(outNum);
//           upper.sum -= outNum;
//           --upper.size;
//         }
//       }
//       // Balance the two maps s.t.
//       // |lower| >= |upper| and |lower| - |upper| <= 1.
//       if (lower.size < upper.size) {
//         final int val = upper.map.firstKey();
//         upper.map.merge(val, -1, Integer::sum);
//         if (upper.map.get(val) == 0)
//           upper.map.remove(val);
//         lower.map.merge(val, 1, Integer::sum);
//         upper.sum -= val;
//         lower.sum += val;
//         --upper.size;
//         ++lower.size;
//       } else if (lower.size - upper.size > 1) {
//         final int val = lower.map.lastKey();
//         lower.map.merge(val, -1, Integer::sum);
//         if (lower.map.get(val) == 0)
//           lower.map.remove(val);
//         upper.map.merge(val, 1, Integer::sum);
//         lower.sum -= val;
//         upper.sum += val;
//         --lower.size;
//         ++upper.size;
//       }
//       // Calculate operations needed to make all elements in the window equal
//       // to the median.
//       if (i >= x - 1) {
//         final int median = lower.map.lastKey();
//         final long ops = (median * lower.size - lower.sum) + (upper.sum - median * upper.size);
//         minOps.add(ops);
//       }
//     }
//     return minOps;
//   }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × m)

Space

O(n × m)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.