Minimum Operations to Exceed Threshold Value II

Minimum Operations to Exceed Threshold Value II - Problem

You are given a 0-indexed integer array nums, and an integer k. You are allowed to perform some operations on nums, where in a single operation, you can:

Select the two smallest integers x and y from nums.
Remove x and y from nums.
Insert (min(x, y) * 2 + max(x, y)) at any position in the array.

Note that you can only apply the described operation if nums contains at least two elements.

Return the minimum number of operations needed so that all elements of the array are greater than or equal to k.

Input & Output

Example 1 — Basic Case

$ Input: nums = [2,11,10,1], k = 10

› Output: 3

💡 Note: Operation 1: combine 1,2 → 1×2+2=4, array=[4,11,10]. Operation 2: combine 4,10 → 4×2+10=18, array=[18,11]. Operation 3: combine 11,18 → 11×2+18=40, array=[40]. All elements ≥ 10, so 3 operations needed.

Example 2 — Already Valid

$ Input: nums = [1,1,2,4,9], k = 1

› Output: 0

💡 Note: All elements are already ≥ 1, so no operations needed.

Example 3 — Single Small Element

$ Input: nums = [1,1,2,4,9], k = 20

› Output: 4

💡 Note: Need multiple operations to combine small elements until all reach threshold 20.

Constraints

2 ≤ nums.length ≤ 10⁵
1 ≤ nums[i] ≤ 10⁶
1 ≤ k ≤ 10⁹

Visualization

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Understanding the Visualization

Input Array

Array [2,11,10,1] with threshold k=10

Combine Process

Always take two smallest: min(x,y)×2 + max(x,y)

Result

Count operations until all elements ≥ k

Key Takeaway

🎯 Key Insight: Use min heap to efficiently find the two smallest elements in O(log n) time

Asked in

G Google 25 f Meta 20 a Amazon 18

The key insight is to always combine the two smallest elements first for optimal growth. The best approach uses a min heap to efficiently access the smallest elements in O(log n) time per operation. Time: O(n log n), Space: O(n).

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force with Array Sorting	O(n² log n)	O(1)	Repeatedly sort array to find smallest elements and combine them
Optimal Solution with Min Heap	O(n log n)	O(n)	Use priority queue to efficiently find the two smallest elements

Brute Force with Array Sorting — Algorithm Steps

Check if all elements >= k
Sort array to find two smallest
Remove smallest two and add their combination
Repeat until done

Visualization

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Step-by-Step Walkthrough

Initial Array

Start with unsorted array

Sort & Combine

Sort entire array, take two smallest, combine them

Repeat

Continue until all elements ≥ k

Code -

solution.c — C

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

int compare(const void *a, const void *b) {
    return (*(int*)a - *(int*)b);
}

int min(int a, int b) {
    return a < b ? a : b;
}

int max(int a, int b) {
    return a > b ? a : b;
}

int solution(int* nums, int numsSize, int k) {
    int* arr = (int*)malloc(numsSize * sizeof(int));
    for (int i = 0; i < numsSize; i++) {
        arr[i] = nums[i];
    }
    
    int operations = 0;
    int currentSize = numsSize;
    
    while (1) {
        // Check if all elements >= k
        int allValid = 1;
        for (int i = 0; i < currentSize; i++) {
            if (arr[i] < k) {
                allValid = 0;
                break;
            }
        }
        
        if (allValid || currentSize < 2) {
            break;
        }
        
        // Sort to find two smallest
        qsort(arr, currentSize, sizeof(int), compare);
        
        // Take two smallest
        int x = arr[0], y = arr[1];
        
        // Remove them and add combination
        for (int i = 0; i < currentSize - 2; i++) {
            arr[i] = arr[i + 2];
        }
        currentSize -= 2;
        
        int combination = min(x, y) * 2 + max(x, y);
        arr[currentSize] = combination;
        currentSize++;
        
        operations++;
    }
    
    free(arr);
    return operations;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    // Parse array manually
    int nums[1000];
    int numsSize = 0;
    char* token = strtok(line, "[],\n");
    while (token != NULL) {
        if (strlen(token) > 0) {
            nums[numsSize++] = atoi(token);
        }
        token = strtok(NULL, "[],\n");
    }
    
    int k;
    scanf("%d", &k);
    
    printf("%d\n", solution(nums, numsSize, k));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n² log n)

Up to n operations, each requiring O(n log n) sorting

⚠ Quadratic Growth

Space Complexity

O(1)

Only using variables for tracking, modifying array in-place

✓ Linear Space

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Medium Frequency

~25 min Avg. Time

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force with Array Sorting — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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