Minimum Operations to Make Numbers Non-positive

Minimum Operations to Make Numbers Non-positive - Problem

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 i-th one

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

Input & Output

Example 1 — Basic Case

$ Input: nums = [3,4,1], x = 3, y = 1

› Output: 2

💡 Note: Operation 1: Choose i=0, nums becomes [0,3,0]. Operation 2: Choose i=1, nums becomes [-1,0,-1]. All numbers ≤ 0 after 2 operations.

Example 2 — Equal x and y

$ Input: nums = [1,2,1], x = 2, y = 2

› Output: 1

💡 Note: Since x = y = 2, each operation decreases all numbers by 2. After 1 operation: nums becomes [-1,0,-1]. All numbers ≤ 0.

Example 3 — Large Difference

$ Input: nums = [10], x = 5, y = 1

› Output: 2

💡 Note: Only one element. Operation 1: nums[0] becomes 5. Operation 2: nums[0] becomes 0. Need 2 operations to make 10 ≤ 0.

Constraints

1 ≤ nums.length ≤ 10⁵
1 ≤ nums[i] ≤ 10⁹
1 ≤ x, y ≤ 10⁹

Visualization

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Asked in

G Google 15 M Meta 12 a Amazon 8 m Microsoft 6

The key insight is to use binary search on the answer space. Instead of trying all operation sequences, we binary search on the number of operations k and check if k operations can make all numbers non-positive. Best approach uses binary search with greedy verification in O(n log(max_val)) time and O(1) space.

Common Approaches

✓ Binary Search on Answer

⏱️ Time: O(n log(max_val)) Space: O(1)

Instead of finding the exact sequence, we binary search on the answer (number of operations k). For each k, we check if it's possible to make all numbers non-positive using exactly k operations.

Brute Force Simulation

⏱️ Time: O(n^k) Space: O(k)

For each operation, try choosing each index and simulate the decrements. Continue until all numbers are <= 0. This explores all possible sequences.

Binary Search on Answer — Algorithm Steps

Step 1: Binary search on operations count k from 0 to max_possible
Step 2: For each k, check feasibility greedily
Step 3: If feasible, try smaller k; otherwise try larger k

Visualization

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Step-by-Step Walkthrough

Search Space

Binary search from 0 to max(nums) operations

Check Feasibility

For each mid value, check if k operations can make all nums ≤ 0

Converge

Narrow down to minimum feasible operations count

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdbool.h>

bool canAchieve(int* nums, int n, int x, int y, long long operations) {
    long long totalNeeded = 0;
    for (int i = 0; i < n; i++) {
        long long target = nums[i] - operations * y;
        if (target <= 0) continue;
        if (x <= y) return false;
        long long needed = (target + x - y - 1) / (x - y);
        totalNeeded += needed;
        if (totalNeeded > operations) return false;
    }
    return true;
}

int solution(int* nums, int n, int x, int y) {
    int left = 0, right = 0;
    for (int i = 0; i < n; i++) {
        if (nums[i] > right) right = nums[i];
    }
    
    while (left < right) {
        int mid = (left + right) / 2;
        if (canAchieve(nums, n, x, y, mid)) {
            right = mid;
        } else {
            left = mid + 1;
        }
    }
    return left;
}

void parseArray(const char* str, int* arr, int* size) {
    *size = 0;
    const char* p = str;
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    while (*p && *p != ']') {
        while (*p == ' ' || *p == ',') p++;
        if (*p == ']' || *p == '\0') break;
        arr[(*size)++] = (int)strtol(p, (char**)&p, 10);
    }
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    int nums[100];
    int n = 0;
    char* token = strtok(line + 1, ",]");
    while (token) {
        nums[n++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    int x, y;
    scanf("%d", &x);
    scanf("%d", &y);
    printf("%d\n", solution(nums, n, x, y));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n log(max_val))

Binary search on answer takes O(log(max_val)) iterations, each checking feasibility in O(n)

✓ Linear Growth

Space Complexity

O(1)

Only using constant extra variables for binary search bounds

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Binary Search on Answer — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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