LeetCode #2111 — HARD

Minimum Operations to Make the Array K-Increasing

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

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The Problem

Problem Statement

You are given a 0-indexed array arr consisting of n positive integers, and a positive integer k.

The array arr is called K-increasing if arr[i-k] <= arr[i] holds for every index i, where k <= i <= n-1.

  • For example, arr = [4, 1, 5, 2, 6, 2] is K-increasing for k = 2 because:
    • arr[0] <= arr[2] (4 <= 5)
    • arr[1] <= arr[3] (1 <= 2)
    • arr[2] <= arr[4] (5 <= 6)
    • arr[3] <= arr[5] (2 <= 2)
  • However, the same arr is not K-increasing for k = 1 (because arr[0] > arr[1]) or k = 3 (because arr[0] > arr[3]).

In one operation, you can choose an index i and change arr[i] into any positive integer.

Return the minimum number of operations required to make the array K-increasing for the given k.

Example 1:

Input: arr = [5,4,3,2,1], k = 1
Output: 4
Explanation:
For k = 1, the resultant array has to be non-decreasing.
Some of the K-increasing arrays that can be formed are [5,6,7,8,9], [1,1,1,1,1], [2,2,3,4,4]. All of them require 4 operations.
It is suboptimal to change the array to, for example, [6,7,8,9,10] because it would take 5 operations.
It can be shown that we cannot make the array K-increasing in less than 4 operations.

Example 2:

Input: arr = [4,1,5,2,6,2], k = 2
Output: 0
Explanation:
This is the same example as the one in the problem description.
Here, for every index i where 2 <= i <= 5, arr[i-2] <= arr[i].
Since the given array is already K-increasing, we do not need to perform any operations.

Example 3:

Input: arr = [4,1,5,2,6,2], k = 3
Output: 2
Explanation:
Indices 3 and 5 are the only ones not satisfying arr[i-3] <= arr[i] for 3 <= i <= 5.
One of the ways we can make the array K-increasing is by changing arr[3] to 4 and arr[5] to 5.
The array will now be [4,1,5,4,6,5].
Note that there can be other ways to make the array K-increasing, but none of them require less than 2 operations.

Constraints:

  • 1 <= arr.length <= 105
  • 1 <= arr[i], k <= arr.length
Patterns Used
🔍Modified Binary Search→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed array arr consisting of n positive integers, and a positive integer k. The array arr is called K-increasing if arr[i-k] <= arr[i] holds for every index i, where k <= i <= n-1. For example, arr = [4, 1, 5, 2, 6, 2] is K-increasing for k = 2 because: arr[0] <= arr[2] (4 <= 5) arr[1] <= arr[3] (1 <= 2) arr[2] <= arr[4] (5 <= 6) arr[3] <= arr[5] (2 <= 2) However, the same arr is not K-increasing for k = 1 (because arr[0] > arr[1]) or k = 3 (because arr[0] > arr[3]). In one operation, you can choose an index i and change arr[i] into any positive integer. Return the minimum number of operations required to make the array K-increasing for the given k.

Baseline thinking

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

Pattern signal: Array · Binary Search

Example 1

[5,4,3,2,1]
1

Example 2

[4,1,5,2,6,2]
2

Example 3

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

Related Problems

  • Longest Increasing Subsequence (longest-increasing-subsequence)
  • Minimum Swaps To Make Sequences Increasing (minimum-swaps-to-make-sequences-increasing)
Step 02

Core Insight

What unlocks the optimal approach

  • Can we divide the array into non-overlapping subsequences and simplify the problem?
  • In the final array, arr[i-k] ≤ arr[i] should hold. We can use this to divide the array into at most k non-overlapping sequences, where arr[i] will belong to the (i%k)th sequence.
  • Now our problem boils down to performing the minimum operations on each sequence such that it becomes non-decreasing. Our answer will be the sum of operations on each sequence.
  • Which indices of a sequence should we not change in order to count the minimum operations? Can finding the longest non-decreasing subsequence of the sequence help?
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. 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 #2111: Minimum Operations to Make the Array K-Increasing
class Solution {
    public int kIncreasing(int[] arr, int k) {
        int n = arr.length;
        int ans = 0;
        for (int i = 0; i < k; ++i) {
            List<Integer> t = new ArrayList<>();
            for (int j = i; j < n; j += k) {
                t.add(arr[j]);
            }
            ans += lis(t);
        }
        return ans;
    }

    private int lis(List<Integer> arr) {
        List<Integer> t = new ArrayList<>();
        for (int x : arr) {
            int idx = searchRight(t, x);
            if (idx == t.size()) {
                t.add(x);
            } else {
                t.set(idx, x);
            }
        }
        return arr.size() - t.size();
    }

    private int searchRight(List<Integer> arr, int x) {
        int left = 0, right = arr.size();
        while (left < right) {
            int mid = (left + right) >> 1;
            if (arr.get(mid) > x) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }
}
Step 06

Interactive Study Demo

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

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(log n)
Space
O(1)

Approach Breakdown

LINEAR SCAN
O(n) time
O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH
O(log n) time
O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.
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.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.

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