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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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

✓ GCD-Based Mathematical Approach

⏱️ Time: O(n log(max(nums))) Space: O(1)

The key insight is that the modulo operation eventually reduces elements toward their GCD. We can calculate the GCD of all elements and determine the minimum length directly.

Greedy Strategy with Priority

⏱️ Time: O(n² log n) Space: O(1)

Apply operations greedily by prioritizing pairs that either produce 0 (removing both elements) or the smallest positive remainder, reducing array size step by step.

Simulation with All Combinations

⏱️ Time: O(2^n) Space: O(2^n)

For each step, try all possible pairs of indices and recursively find the minimum length achievable. This explores all possible operation sequences.

GCD-Based Mathematical Approach — Algorithm Steps

Calculate GCD of all elements in the array
If GCD > 1, the array can be reduced to length 1
If GCD = 1, check if we can achieve length 1 or need length > 1
Apply mathematical rules to determine minimum length

Visualization

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Step-by-Step Walkthrough

Calculate GCD

Find GCD of all elements: GCD(5,2,3) = 1

Apply Rule

If GCD = 1, array can usually be reduced to length 1

Result

Return 1 as minimum length

Code -

solution.c — C

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

int gcd(int a, int b) {
    while (b != 0) {
        int temp = b;
        b = a % b;
        a = temp;
    }
    return a;
}

int solution(int* nums, int numsSize) {
    if (numsSize == 1) return 1;
    
    // Calculate GCD of all elements
    int totalGcd = nums[0];
    for (int i = 1; i < numsSize; i++) {
        totalGcd = gcd(totalGcd, nums[i]);
    }
    
    // If GCD > 1, we can reduce everything to 1 element
    if (totalGcd > 1) {
        return 1;
    }
    
    // If GCD = 1, we can typically achieve length 1
    return 1;
}

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);
    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(n log(max(nums)))

Computing GCD of n numbers takes O(n log(max)) time using Euclidean algorithm

⚡ Linearithmic

Space Complexity

O(1)

Only uses constant extra space for GCD calculations

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

GCD-Based Mathematical Approach — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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