LeetCode #3375 — EASY

Minimum Operations to Make Array Values Equal to K

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

You are given an integer array nums and an integer k.

An integer h is called valid if all values in the array that are strictly greater than h are identical.

For example, if nums = [10, 8, 10, 8], a valid integer is h = 9 because all nums[i] > 9 are equal to 10, but 5 is not a valid integer.

You are allowed to perform the following operation on nums:

Select an integer h that is valid for the current values in nums.
For each index i where nums[i] > h, set nums[i] to h.

Return the minimum number of operations required to make every element in nums equal to k. If it is impossible to make all elements equal to k, return -1.

Example 1:

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

Output: 2

Explanation:

The operations can be performed in order using valid integers 4 and then 2.

Example 2:

Input: nums = [2,1,2], k = 2

Output: -1

Explanation:

It is impossible to make all the values equal to 2.

Example 3:

Input: nums = [9,7,5,3], k = 1

Output: 4

Explanation:

The operations can be performed using valid integers in the order 7, 5, 3, and 1.

Constraints:

1 <= nums.length <= 100
1 <= nums[i] <= 100
1 <= k <= 100

Step 01

Brute Force Baseline

Problem summary: You are given an integer array nums and an integer k. An integer h is called valid if all values in the array that are strictly greater than h are identical. For example, if nums = [10, 8, 10, 8], a valid integer is h = 9 because all nums[i] > 9 are equal to 10, but 5 is not a valid integer. You are allowed to perform the following operation on nums: Select an integer h that is valid for the current values in nums. For each index i where nums[i] > h, set nums[i] to h. Return the minimum number of operations required to make every element in nums equal to k. If it is impossible to make all elements equal to k, return -1.

Baseline thinking

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

Pattern signal: Array · Hash Map

Example 1

[5,2,5,4,5]
2

Example 2

[2,1,2]
2

Example 3

[9,7,5,3]
1

Step 02

Core Insight

What unlocks the optimal approach

Handle the case when the array contains an integer less than <code>k</code>
Start by performing operations on the highest integer
You can perform an operation on the highest integer using the second-highest, an operation on the second-highest using the third-highest, and so forth.
The answer is the number of distinct integers in the array that are larger than <code>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

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 #3375: Minimum Operations to Make Array Values Equal to K
class Solution {
    public int minOperations(int[] nums, int k) {
        Set<Integer> s = new HashSet<>();
        int mi = 1 << 30;
        for (int x : nums) {
            if (x < k) {
                return -1;
            }
            mi = Math.min(mi, x);
            s.add(x);
        }
        return s.size() - (mi == k ? 1 : 0);
    }
}

// Accepted solution for LeetCode #3375: Minimum Operations to Make Array Values Equal to K
func minOperations(nums []int, k int) int {
	mi := 1 << 30
	s := map[int]bool{}
	for _, x := range nums {
		if x < k {
			return -1
		}
		s[x] = true
		mi = min(mi, x)
	}
	if mi == k {
		return len(s) - 1
	}
	return len(s)
}

# Accepted solution for LeetCode #3375: Minimum Operations to Make Array Values Equal to K
class Solution:
    def minOperations(self, nums: List[int], k: int) -> int:
        s = set()
        mi = inf
        for x in nums:
            if x < k:
                return -1
            mi = min(mi, x)
            s.add(x)
        return len(s) - int(k == mi)

// Accepted solution for LeetCode #3375: Minimum Operations to Make Array Values Equal to K
impl Solution {
    pub fn min_operations(nums: Vec<i32>, k: i32) -> i32 {
        use std::collections::HashSet;

        let mut s = HashSet::new();
        let mut mi = i32::MAX;

        for &x in &nums {
            if x < k {
                return -1;
            }
            s.insert(x);
            mi = mi.min(x);
        }

        (s.len() as i32) - if mi == k { 1 } else { 0 }
    }
}

// Accepted solution for LeetCode #3375: Minimum Operations to Make Array Values Equal to K
function minOperations(nums: number[], k: number): number {
    const s = new Set<number>([k]);
    for (const x of nums) {
        if (x < k) return -1;
        s.add(x);
    }
    return s.size - 1;
}

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)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

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.