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:
h that is valid for the current values in nums.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] <= 1001 <= k <= 100Problem Overview: You are given an integer array nums and a target value k. The goal is to determine the minimum number of operations needed so every element becomes exactly k. An operation effectively eliminates a value greater than k by converting all occurrences of that value to k. If any number is already smaller than k, the task becomes impossible because operations only reduce larger values to k.
Approach 1: Hash Set to Track Distinct Values ( O(n) time, O(n) space )
Start by scanning the array once. If you encounter any value smaller than k, immediately return -1 because no valid operation can increase values. For values greater than k, insert them into a set (or hash table). Each distinct value greater than k represents one required operation, since the operation reduces that specific value to k for all its occurrences. After processing the array, the answer equals the number of unique elements stored in the set.
The key insight: multiple occurrences of the same value can be handled in a single operation. Only the number of distinct values above k matters. A hash table or set provides constant‑time insertion and membership checks, making the solution linear.
Implementation is straightforward: iterate through the array, validate values relative to k, and track distinct values larger than k. Languages like Python, Java, and C++ all provide efficient built‑in hash set structures.
Recommended for interviews: The hash‑set approach is the expected solution. Interviewers want to see that you recognize two constraints quickly: values smaller than k make the task impossible, and duplicate values above k should only count once. Explaining this observation shows strong reasoning about hash-based deduplication and linear scans. Brute‑force counting without deduplication leads to incorrect results, so identifying the need for a set demonstrates the key insight.
According to the problem description, we can choose the second largest value in the current array as the valid integer h each time, and change all numbers greater than h to h. This minimizes the number of operations. Additionally, since the operation reduces the numbers, if there are numbers in the current array smaller than k, we cannot make all numbers equal to k, so we directly return -1.
We iterate through the array nums. For the current number x, if x < k, we directly return -1. Otherwise, we add x to the hash table and update the minimum value mi in the current array. Finally, we return the size of the hash table minus 1 (if mi = k, we need to subtract 1).
Time complexity is O(n), and space complexity is O(n), where n is the length of the array nums.
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| Approach | Time | Space | When to Use |
|---|---|---|---|
| Hash Set (Count Distinct Values > k) | O(n) | O(n) | General case. Efficient for unsorted arrays and handles duplicates naturally. |
| Sorting + Unique Scan | O(n log n) | O(1) or O(n) | Useful when the array is already sorted or when avoiding hash structures. |
Minimum Operations to Make Array Values Equal to K | Detailed Thought Process | Leetcode 3375 | MIK • codestorywithMIK • 6,444 views views
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