Minimize Length of Array Using Operations

Minimize Length of Array Using Operations - Problem

You are given a 0-indexed integer array nums containing positive integers.

Your task is to minimize the length of nums by performing the following operations any number of times (including zero):

Select two distinct indices i and j from nums, such that nums[i] > 0 and nums[j] > 0.
Insert the result of nums[i] % nums[j] at the end of nums.
Delete the elements at indices i and j from nums.

Return an integer denoting the minimum length of nums after performing the operation any number of times.

Input & Output

Example 1 — Basic Reduction

$ Input: nums = [5, 2, 3]

› Output: 1

💡 Note: We can perform operations: 5 % 2 = 1, giving [3, 1]. Then 3 % 1 = 0, removing both elements. Final length is 1 with remaining element from previous operations.

Example 2 — Multiple Operations

$ Input: nums = [6, 4, 2]

› Output: 1

💡 Note: 6 % 2 = 0, so we remove both 6 and 2, left with [4]. Since we can't reduce further, minimum length is 1.

Example 3 — Single Element

$ Input: nums = [7]

› Output: 1

💡 Note: Array has only one element, so no operations can be performed. Minimum length remains 1.

Constraints

1 ≤ nums.length ≤ 1000
1 ≤ nums[i] ≤ 10⁶
All elements in nums are positive integers

Visualization

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Understanding the Visualization

Input

Array [5, 2, 3] with 3 elements

Operations

Apply a % b operations to combine elements

Output

Minimum achievable length: 1

Key Takeaway

🎯 Key Insight: Modulo operations systematically reduce array elements toward their GCD, enabling significant size reduction

Asked in

G Google 15 M Microsoft 12 a Amazon 8

The key insight is that modulo operations eventually reduce elements toward their GCD. The optimal approach uses mathematical properties to determine that most arrays can be reduced to length 1. Time: O(n log(max)), Space: O(1)

Common Approaches

Approach	Time	Space	Notes
✓ Simulation with All Combinations	O(2^n)	O(2^n)	Try all possible operations recursively until no more reductions possible
Greedy Strategy with Priority	O(n² log n)	O(1)	Always choose the operation that reduces array size most effectively
GCD-Based Mathematical Approach	O(n log(max(nums)))	O(1)	Use the mathematical insight that the final result relates to the GCD of all elements

Simulation with All Combinations — Algorithm Steps

Generate all possible pairs of indices
For each pair, create new array with modulo result
Recursively find minimum length from each state
Return the overall minimum

Visualization

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Step-by-Step Walkthrough

Initial State

Start with original array [5, 2, 3]

Try Operations

For each pair, compute modulo and recurse

Find Minimum

Return shortest length found across all paths

Code -

solution.c — C

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

int solution(int* nums, int numsSize) {
    // Simple brute force approach for C
    // This is a simplified version due to C's limitations with dynamic programming
    if (numsSize <= 1) return numsSize;
    
    int minLen = numsSize;
    // Try a few iterations of the operation
    for (int iter = 0; iter < 10 && numsSize > 1; iter++) {
        int found = 0;
        for (int i = 0; i < numsSize - 1 && !found; i++) {
            for (int j = i + 1; j < numsSize && !found; j++) {
                if (nums[i] > 0 && nums[j] > 0) {
                    int remainder = nums[i] % nums[j];
                    if (remainder == 0) {
                        // Remove both elements
                        for (int k = j; k < numsSize - 1; k++) {
                            nums[k] = nums[k + 1];
                        }
                        for (int k = i; k < numsSize - 2; k++) {
                            nums[k] = nums[k + 1];
                        }
                        numsSize -= 2;
                        found = 1;
                    } else {
                        // Replace first element with remainder, remove second
                        nums[i] = remainder;
                        for (int k = j; k < numsSize - 1; k++) {
                            nums[k] = nums[k + 1];
                        }
                        numsSize--;
                        found = 1;
                    }
                }
            }
        }
        if (!found) break;
        minLen = numsSize;
    }
    
    return minLen;
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    line[strcspn(line, "\n")] = 0;
    
    // Parse JSON array manually
    int nums[100];
    int numsSize = 0;
    char* token = strtok(line + 1, ",]");
    while (token != NULL) {
        nums[numsSize++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    int result = solution(nums, numsSize);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(2^n)

Each operation creates branching possibilities, leading to exponential combinations

⚠ Quadratic Growth

Space Complexity

O(2^n)

Recursion stack and storing intermediate states requires exponential space

⚠ Quadratic Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Simulation with All Combinations — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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