2702. Minimum Operations to Make Numbers Non-positive 🔒

Description

You are given a 0-indexed integer array nums and two integers x and y. In one operation, you must choose an index i such that 0 <= i < nums.length and perform the following:

• Decrement nums[i] by x.
• Decrement values by y at all indices except the ith one.

Return the minimum number of operations to make all the integers in nums less than or equal to zero.

 

Example 1:

Input: nums = [3,4,1,7,6], x = 4, y = 2
Output: 3
Explanation: You will need three operations. One of the optimal sequence of operations is:
Operation 1: Choose i = 3. Then, nums = [1,2,-1,3,4]. 
Operation 2: Choose i = 3. Then, nums = [-1,0,-3,-1,2].
Operation 3: Choose i = 4. Then, nums = [-3,-2,-5,-3,-2].
Now, all the numbers in nums are non-positive. Therefore, we return 3.

Example 2:

Input: nums = [1,2,1], x = 2, y = 1
Output: 1
Explanation: We can perform the operation once on i = 1. Then, nums becomes [0,0,0]. All the positive numbers are removed, and therefore, we return 1.

 

Constraints:

• 1 <= nums.length <= 105
• 1 <= nums[i] <= 109
• 1 <= y < x <= 109

Solutions

Solution 1: Binary Search

We notice that if an operation count $t$ can make all numbers less than or equal to $0$, then for any $t' > t$, the operation count $t'$ can also make all numbers less than or equal to $0$. Therefore, we can use binary search to find the minimum operation count.

We define the left boundary of the binary search as $l=0$, and the right boundary as $r=\max(nums)$. Each time we perform a binary search, we find the middle value $mid=\lfloor\frac{l+r}{2}\rfloor$, and then determine whether there exists an operation method that does not exceed $mid$ and makes all numbers less than or equal to $0$. If it exists, we update the right boundary $r = mid$, otherwise, we update the left boundary

Python3JavaC++GoTypeScript

class Solution:
    def minOperations(self, nums: List[int], x: int, y: int) -> int:
        def check(t: int) -> bool:
            cnt = 0
            for v in nums:
                if v > t * y:
                    cnt += ceil((v - t * y) / (x - y))
            return cnt <= t

        l, r = 0, max(nums)
        while l < r:
            mid = (l + r) >> 1
            if check(mid):
                r = mid
            else:
                l = mid + 1
        return l

class Solution {
    private int[] nums;
    private int x;
    private int y;

    public int minOperations(int[] nums, int x, int y) {
        this.nums = nums;
        this.x = x;
        this.y = y;
        int l = 0, r = 0;
        for (int v : nums) {
            r = Math.max(r, v);
        }
        while (l < r) {
            int mid = (l + r) >>> 1;
            if (check(mid)) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    }

    private boolean check(int t) {
        long cnt = 0;
        for (int v : nums) {
            if (v > (long) t * y) {
                cnt += (v - (long) t * y + x - y - 1) / (x - y);
            }
        }
        return cnt <= t;
    }
}

class Solution {
public:
    int minOperations(vector<int>& nums, int x, int y) {
        int l = 0, r = *max_element(nums.begin(), nums.end());
        auto check = [&](int t) {
            long long cnt = 0;
            for (int v : nums) {
                if (v > 1LL * t * y) {
                    cnt += (v - 1LL * t * y + x - y - 1) / (x - y);
                }
            }
            return cnt <= t;
        };
        while (l < r) {
            int mid = (l + r) >> 1;
            if (check(mid)) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    }
};

func minOperations(nums []int, x int, y int) int {
    check := func(t int) bool {
        cnt := 0
        for _, v := range nums {
            if v > t*y {
                cnt += (v - t*y + x - y - 1) / (x - y)
            }
        }
        return cnt <= t
    }

    l, r := 0, slices.Max(nums)
    for l < r {
        mid := (l + r) >> 1
        if check(mid) {
            r = mid
        } else {
            l = mid + 1
        }
    }
    return l
}

function minOperations(nums: number[], x: number, y: number): number {
    let l = 0;
    let r = Math.max(...nums);
    const check = (t: number): boolean => {
        let cnt = 0;
        for (const v of nums) {
            if (v > t * y) {
                cnt += Math.ceil((v - t * y) / (x - y));
            }
        }
        return cnt <= t;
    };
    while (l < r) {
        const mid = (l + r) >> 1;
        if (check(mid)) {
            r = mid;
        } else {
            l = mid + 1;
        }
    }
    return l;
}

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