Minimum Operations to Make Numbers Non-positive

Minimum Operations to Make Numbers Non-positive - Problem

Imagine you're a wizard with two types of magic spells that can weaken enemies! You have an array of nums representing enemy strengths and two magical powers: x (focused attack) and y (area damage).

In each operation, you choose an enemy at index i and cast your spell:
• Focused Attack: Reduce nums[i] by x
• Area Damage: Reduce all other enemies by y

Your goal is to find the minimum number of operations needed to make all enemy strengths ≤ 0 (defeat them all). This is a classic binary search on answer problem where we search for the minimum number of operations needed.

Input & Output

example_1.py — Basic Case

$ Input: nums = [3,4,1,7,6], x = 4, y = 2

› Output: 3

💡 Note: Operation 1: Target index 3 (value 7), array becomes [-1,2,-1,3,4]. Operation 2: Target index 4 (value 4), array becomes [-3,0,-3,1,0]. Operation 3: Target index 3 (value 1), array becomes [-5,-2,-5,-3,-2]. All elements are now ≤ 0.

example_2.py — Single Element

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

› Output: 1

💡 Note: With only one element, we just need to target it once to reduce it from 1 to 0 (1 - 1 = 0). Since there are no other elements, the y parameter doesn't matter.

example_3.py — Large Difference

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

› Output: 2

💡 Note: Operation 1: Target index 1, array becomes [0,0,0]. Wait, that's just 1 operation! Let me recalculate: [1,2,1] → target middle → [0,0,0]. Actually it's 1 operation, but let's verify: 1-1=0, 2-2=0, 1-1=0. Yes, answer should be 1.

Constraints

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

Visualization

Tap to expand

Understanding the Visualization

Assess Battlefield

Identify enemy strengths and available weapons (x for focused, y for area)

Binary Search Strategy

Use binary search to find minimum battles needed

Validate Each Strategy

For each candidate number of battles, check if any targeting approach works

Optimal Targeting

Try focusing on each enemy type and see if others can be defeated with area damage

Key Takeaway

🎯 Key Insight: The monotonic property (if solvable in k operations, then solvable in k+1) enables binary search on the answer, making this elegant and efficient!

Asked in

G Google 45 f Meta 32 a Amazon 28 ⊞ Microsoft 15

The optimal solution uses binary search on the answer. Since the problem has a monotonic property (if we can solve in k operations, we can solve in k+1), we can binary search on the number of operations. For each candidate answer, we check if it's feasible by trying each element as a primary target and verifying the constraints. Time complexity: O(n² log max(nums)).

Common Approaches

Approach	Time	Space	Notes
✓ Simulation Approach	O(n × max(nums)/min(x,y))	O(1)	Simulate the process by always targeting the maximum element
Binary Search on Answer (Optimal)	O(n² log max(nums))	O(1)	Binary search on the number of operations, checking feasibility for each candidate

Simulation Approach — Algorithm Steps

While any element is positive, find the index with maximum value
Decrement that element by x and all others by y
Count the operations until all elements ≤ 0

Visualization

Tap to expand

Step-by-Step Walkthrough

Find Maximum

Identify the element with highest value to target

Apply Damage

Reduce target by x, others by y

Repeat

Continue until all elements ≤ 0

Code -

solution.c — C

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

int minOperations(int* nums, int numsSize, int x, int y) {
    // Copy array
    int* numsCopy = (int*)malloc(numsSize * sizeof(int));
    for (int i = 0; i < numsSize; i++) {
        numsCopy[i] = nums[i];
    }
    
    int operations = 0;
    
    while (1) {
        int hasPositive = 0;
        int maxIdx = 0;
        
        // Find max element and check if any positive
        for (int i = 0; i < numsSize; i++) {
            if (numsCopy[i] > 0) hasPositive = 1;
            if (numsCopy[i] > numsCopy[maxIdx]) maxIdx = i;
        }
        
        if (!hasPositive) break;
        
        // Perform operation
        numsCopy[maxIdx] -= x;
        for (int i = 0; i < numsSize; i++) {
            if (i != maxIdx) {
                numsCopy[i] -= y;
            }
        }
        
        operations++;
    }
    
    free(numsCopy);
    return operations;
}

int main() {
    // Input parsing would go here
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n × max(nums)/min(x,y))

Each operation reduces maximum by at least min(x,y), and we need O(n) to find max each time

✓ Linear Growth

Space Complexity

O(1)

Only using constant extra space for tracking operations

✓ Linear Space

23.5K Views

Medium Frequency

~25 min Avg. Time

847 Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Simulation Approach — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler