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3375. Minimum Operations to Make Array Values Equal to K

Description

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

Solutions

Solution 1

  • 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);
    }
}


  • class Solution {
public:
    int minOperations(vector<int>& nums, int k) {
        unordered_set<int> s;
        int mi = INT_MAX;
        for (int x : nums) {
            if (x < k) {
                return -1;
            }
            mi = min(mi, x);
            s.insert(x);
        }
        return s.size() - (mi == 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)


  • 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)
}


  • function minOperations(nums: number[], k: number): number {
    const s = new Set<number>();
    let mi = Infinity;
    for (const x of nums) {
        if (x < k) {
            return -1;
        }
        s.add(x);
        mi = Math.min(mi, x);
    }
    return s.size - (mi === k ? 1 : 0);
}

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