Apply Operations to Make Sum of Array Greater Than or Equal to k - Problem

You are given a positive integer k. Initially, you have an array nums = [1].

You can perform any of the following operations on the array any number of times (possibly zero):

Choose any element in the array and increase its value by 1.
Duplicate any element in the array and add it to the end of the array.

Return the minimum number of operations required to make the sum of elements of the final array greater than or equal to k.

Input & Output

Example 1 — Small Target

$ Input: k = 4

› Output: 2

💡 Note: Start with [1]. Duplicate to get [1,1] (1 op). Duplicate again to get [1,1,1,1] with sum=4 (total 2 ops).

Example 2 — Need Increments

$ Input: k = 11

› Output: 6

💡 Note: Start with [1]. After 3 duplicates: [1,1,1,1,1,1,1,1] sum=8. Then 3 increments to reach sum=11. Total: 3+3=6 operations.

Example 3 — Already Satisfied

$ Input: k = 1

› Output: 0

💡 Note: Initial array [1] already has sum=1 ≥ k=1, so no operations needed.

Constraints

1 ≤ k ≤ 10⁵

Visualization

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Asked in

G Google 15 f Meta 12 a Amazon 8

The key insight is to find the optimal balance between duplicating (doubling array size) and incrementing elements. The best approach uses mathematical enumeration to try different numbers of duplicate operations and find the minimum total cost. Time: O(log k), Space: O(1)

Common Approaches

✓ Brute Force Enumeration

⏱️ Time: O(2^k) Space: O(k)

Generate all possible sequences of operations (increment vs duplicate) and find the one that reaches sum >= k with minimum operations.

Greedy Strategy

⏱️ Time: O(log k) Space: O(1)

At each step, choose the operation that gives maximum sum increase per operation cost. Duplicate when array is small, increment when array is large.

Optimal Mathematical Solution

⏱️ Time: O(log k) Space: O(1)

Analyze when to stop duplicating and start incrementing. The optimal strategy is to duplicate until the next duplicate would overshoot significantly, then increment the remainder.

Brute Force Enumeration — Algorithm Steps

Generate all possible operation sequences
For each sequence, calculate final sum
Return minimum operations among valid sequences

Visualization

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Step-by-Step Walkthrough

Initial State

Start with array [1], sum = 1

Branch Operations

Try both increment and duplicate at each step

Find Minimum

Return shortest path to sum >= k

Code -

solution.c — C

#include <stdio.h>
#include <limits.h>

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

int dfs(int currentSum, int arraySize, int ops, int k) {
    if (currentSum >= k) {
        return ops;
    }
    
    if (ops > 20) {
        return INT_MAX;
    }
    
    int minOps = INT_MAX;
    
    minOps = min(minOps, dfs(currentSum + 1, arraySize, ops + 1, k));
    minOps = min(minOps, dfs(currentSum * 2, arraySize * 2, ops + 1, k));
    
    return minOps;
}

int solution(int k) {
    return dfs(1, 1, 0, k);
}

int main() {
    int k;
    scanf("%d", &k);
    
    int result = solution(k);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(2^k)

Each operation has 2 choices, leading to exponential combinations

⚡ Linearithmic

Space Complexity

O(k)

Recursion stack depth proportional to number of operations

✓ Linear Space

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Apply Operations to Make Sum of Array Greater Than or Equal to k - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force Enumeration — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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