LeetCode #3347 — HARD

Maximum Frequency of an Element After Performing Operations II

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 k and numOperations.

You must perform an operation numOperations times on nums, where in each operation you:

Select an index i that was not selected in any previous operations.
Add an integer in the range [-k, k] to nums[i].

Return the maximum possible frequency of any element in nums after performing the operations.

Example 1:

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

Output: 2

Explanation:

We can achieve a maximum frequency of two by:

Adding 0 to nums[1], after which nums becomes [1, 4, 5].
Adding -1 to nums[2], after which nums becomes [1, 4, 4].

Example 2:

Input: nums = [5,11,20,20], k = 5, numOperations = 1

Output: 2

Explanation:

We can achieve a maximum frequency of two by:

Adding 0 to nums[1].

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁹
0 <= k <= 10⁹
0 <= numOperations <= nums.length

Step 01

Brute Force Baseline

Problem summary: You are given an integer array nums and two integers k and numOperations. You must perform an operation numOperations times on nums, where in each operation you: Select an index i that was not selected in any previous operations. Add an integer in the range [-k, k] to nums[i]. Return the maximum possible frequency of any element in nums after performing the operations.

Baseline thinking

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

Pattern signal: Array · Binary Search · Sliding Window

Example 1

[1,4,5]
1
2

Example 2

[5,11,20,20]
5
1

Core Insight

What unlocks the optimal approach

The optimal values to check are <code>nums[i] - k</code>, <code>nums[i]</code>, and <code>nums[i] + k</code>.

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 #3347: Maximum Frequency of an Element After Performing Operations II
class Solution {
    public int maxFrequency(int[] nums, int k, int numOperations) {
        Map<Integer, Integer> cnt = new HashMap<>();
        TreeMap<Integer, Integer> d = new TreeMap<>();
        for (int x : nums) {
            cnt.merge(x, 1, Integer::sum);
            d.putIfAbsent(x, 0);
            d.merge(x - k, 1, Integer::sum);
            d.merge(x + k + 1, -1, Integer::sum);
        }
        int ans = 0, s = 0;
        for (var e : d.entrySet()) {
            int x = e.getKey(), t = e.getValue();
            s += t;
            ans = Math.max(ans, Math.min(s, cnt.getOrDefault(x, 0) + numOperations));
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3347: Maximum Frequency of an Element After Performing Operations II
func maxFrequency(nums []int, k int, numOperations int) (ans int) {
	cnt := make(map[int]int)
	d := make(map[int]int)
	for _, x := range nums {
		cnt[x]++
		d[x] = d[x]
		d[x-k]++
		d[x+k+1]--
	}

	s := 0
	keys := make([]int, 0, len(d))
	for key := range d {
		keys = append(keys, key)
	}
	sort.Ints(keys)
	for _, x := range keys {
		s += d[x]
		ans = max(ans, min(s, cnt[x]+numOperations))
	}

	return
}

# Accepted solution for LeetCode #3347: Maximum Frequency of an Element After Performing Operations II
class Solution:
    def maxFrequency(self, nums: List[int], k: int, numOperations: int) -> int:
        cnt = defaultdict(int)
        d = defaultdict(int)
        for x in nums:
            cnt[x] += 1
            d[x] += 0
            d[x - k] += 1
            d[x + k + 1] -= 1
        ans = s = 0
        for x, t in sorted(d.items()):
            s += t
            ans = max(ans, min(s, cnt[x] + numOperations))
        return ans

// Accepted solution for LeetCode #3347: Maximum Frequency of an Element After Performing Operations II
use std::collections::{HashMap, BTreeMap};

impl Solution {
    pub fn max_frequency(nums: Vec<i32>, k: i32, num_operations: i32) -> i32 {
        let mut cnt = HashMap::new();
        let mut d = BTreeMap::new();

        for &x in &nums {
            *cnt.entry(x).or_insert(0) += 1;
            d.entry(x).or_insert(0);
            *d.entry(x - k).or_insert(0) += 1;
            *d.entry(x + k + 1).or_insert(0) -= 1;
        }

        let mut ans = 0;
        let mut s = 0;
        for (&x, &t) in d.iter() {
            s += t;
            let cur = s.min(cnt.get(&x).copied().unwrap_or(0) + num_operations);
            ans = ans.max(cur);
        }

        ans
    }
}

// Accepted solution for LeetCode #3347: Maximum Frequency of an Element After Performing Operations II
function maxFrequency(nums: number[], k: number, numOperations: number): number {
    const cnt: Record<number, number> = {};
    const d: Record<number, number> = {};
    for (const x of nums) {
        cnt[x] = (cnt[x] || 0) + 1;
        d[x] = d[x] || 0;
        d[x - k] = (d[x - k] || 0) + 1;
        d[x + k + 1] = (d[x + k + 1] || 0) - 1;
    }
    let [ans, s] = [0, 0];
    const keys = Object.keys(d)
        .map(Number)
        .sort((a, b) => a - b);
    for (const x of keys) {
        s += d[x];
        ans = Math.max(ans, Math.min(s, (cnt[x] || 0) + numOperations));
    }

    return ans;
}

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 × log n)

Space

O(n)

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.

Shrinking the window only once

Wrong move: Using `if` instead of `while` leaves the window invalid for multiple iterations.

Usually fails on: Over-limit windows stay invalid and produce wrong lengths/counts.

Fix: Shrink in a `while` loop until the invariant is valid again.