LeetCode #3434 — MEDIUM

Maximum Frequency After Subarray Operation

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode

The Problem

Problem Statement

You are given an array nums of length n. You are also given an integer k.

You perform the following operation on nums once:

Select a subarray nums[i..j] where 0 <= i <= j <= n - 1.
Select an integer x and add x to all the elements in nums[i..j].

Find the maximum frequency of the value k after the operation.

Example 1:

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

Output: 2

Explanation:

After adding -5 to nums[2..5], 1 has a frequency of 2 in [1, 2, -2, -1, 0, 1].

Example 2:

Input: nums = [10,2,3,4,5,5,4,3,2,2], k = 10

Output: 4

Explanation:

After adding 8 to nums[1..9], 10 has a frequency of 4 in [10, 10, 11, 12, 13, 13, 12, 11, 10, 10].

Constraints:

1 <= n == nums.length <= 10⁵
1 <= nums[i] <= 50
1 <= k <= 50

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 array nums of length n. You are also given an integer k. You perform the following operation on nums once: Select a subarray nums[i..j] where 0 <= i <= j <= n - 1. Select an integer x and add x to all the elements in nums[i..j]. Find the maximum frequency of the value k after the operation.

Baseline thinking

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

Pattern signal: Array · Hash Map · Dynamic Programming · Greedy

Example 1

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

Example 2

[10,2,3,4,5,5,4,3,2,2]
10

Step 02

Core Insight

What unlocks the optimal approach

Fix the element you want to convert to <code>k</code>.
Use prefix sums to optimize counting occurrences of an element.

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

Upper-end input sizes

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 #3434: Maximum Frequency After Subarray Operation
class Solution {
  public int maxFrequency(int[] nums, int k) {
    final int MAX = 50;
    int maxFreq = 0;
    for (int target = 1; target <= MAX; ++target)
      if (target != k)
        maxFreq = Math.max(maxFreq, kadane(nums, target, k));
    return (int) Arrays.stream(nums).filter(num -> num == k).count() + maxFreq;
  }

  // Returns the maximum achievable frequency of `k` using Kakane's algorithm,
  // where each `target` in subarrays is transformed to `k`.
  private int kadane(int[] nums, int target, int k) {
    int maxSum = 0;
    int sum = 0;
    for (final int num : nums) {
      if (num == target)
        ++sum;
      else if (num == k)
        --sum;
      if (sum < 0) // Reset if sum becomes negative (Kadane's spirit).
        sum = 0;
      maxSum = Math.max(maxSum, sum);
    }

    return maxSum;
  }
}

// Accepted solution for LeetCode #3434: Maximum Frequency After Subarray Operation
package main

func maxFrequency(nums []int, k int) int {
	f0, maxF12 := 0, 0
	f1 := [51]int{}
	for _, x := range nums {
		if x == k {
			maxF12++
			f0++
		} else {
			f1[x] = max(f1[x], f0) + 1
			maxF12 = max(maxF12, f1[x])
		}
	}
	return maxF12
}

func maxFrequency2(nums []int, k int) int {
	var f0, maxF1, f2 int
	f1 := [51]int{}
	for _, x := range nums {
		f2 = max(f2, maxF1)
		f1[x] = max(f1[x], f0) + 1
		if x == k {
			f2++
			f0++
		}
		maxF1 = max(maxF1, f1[x])
	}
	return max(maxF1, f2)
}

func maxFrequency1(nums []int, k int) (ans int) {
	set := map[int]struct{}{}
	for _, x := range nums {
		set[x] = struct{}{}
	}

	for target := range set {
		var f0, f1, f2 int
		for _, x := range nums {
			f2 = max(f2, f1) + b2i(x == k)
			f1 = max(f1, f0) + b2i(x == target)
			f0 += b2i(x == k)
		}
		ans = max(ans, f1, f2)
	}
	return
}

func b2i(b bool) int {
	if b {
		return 1
	}
	return 0
}

# Accepted solution for LeetCode #3434: Maximum Frequency After Subarray Operation
class Solution:
  def maxFrequency(self, nums: list[int], k: int) -> int:
    return nums.count(k) + max(self._kadane(nums, target, k)
                               for target in range(1, 51)
                               if target != k)

  def _kadane(self, nums: list[int], target: int, k: int) -> int:
    """
    Returns the maximum achievable frequency of `k` by Kakane's algorithm,
    where each `target` in subarrays is transformed to `k`.
    """
    maxSum = 0
    sum = 0
    for num in nums:
      if num == target:
        sum += 1
      elif num == k:
        sum -= 1
      if sum < 0:  # Reset sum if it becomes negative (Kadane's spirit).
        sum = 0
      maxSum = max(maxSum, sum)
    return maxSum

// Accepted solution for LeetCode #3434: Maximum Frequency After Subarray Operation
// 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 #3434: Maximum Frequency After Subarray Operation
// class Solution {
//   public int maxFrequency(int[] nums, int k) {
//     final int MAX = 50;
//     int maxFreq = 0;
//     for (int target = 1; target <= MAX; ++target)
//       if (target != k)
//         maxFreq = Math.max(maxFreq, kadane(nums, target, k));
//     return (int) Arrays.stream(nums).filter(num -> num == k).count() + maxFreq;
//   }
// 
//   // Returns the maximum achievable frequency of `k` using Kakane's algorithm,
//   // where each `target` in subarrays is transformed to `k`.
//   private int kadane(int[] nums, int target, int k) {
//     int maxSum = 0;
//     int sum = 0;
//     for (final int num : nums) {
//       if (num == target)
//         ++sum;
//       else if (num == k)
//         --sum;
//       if (sum < 0) // Reset if sum becomes negative (Kadane's spirit).
//         sum = 0;
//       maxSum = Math.max(maxSum, sum);
//     }
// 
//     return maxSum;
//   }
// }

// Accepted solution for LeetCode #3434: Maximum Frequency After Subarray Operation
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #3434: Maximum Frequency After Subarray Operation
// class Solution {
//   public int maxFrequency(int[] nums, int k) {
//     final int MAX = 50;
//     int maxFreq = 0;
//     for (int target = 1; target <= MAX; ++target)
//       if (target != k)
//         maxFreq = Math.max(maxFreq, kadane(nums, target, k));
//     return (int) Arrays.stream(nums).filter(num -> num == k).count() + maxFreq;
//   }
// 
//   // Returns the maximum achievable frequency of `k` using Kakane's algorithm,
//   // where each `target` in subarrays is transformed to `k`.
//   private int kadane(int[] nums, int target, int k) {
//     int maxSum = 0;
//     int sum = 0;
//     for (final int num : nums) {
//       if (num == target)
//         ++sum;
//       else if (num == k)
//         --sum;
//       if (sum < 0) // Reset if sum becomes negative (Kadane's spirit).
//         sum = 0;
//       maxSum = Math.max(maxSum, sum);
//     }
// 
//     return maxSum;
//   }
// }

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.

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.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.