Minimum Split Into Subarrays With GCD Greater Than One

Minimum Split Into Subarrays With GCD Greater Than One - Problem

Array Math Dynamic Programming Greedy Number Theory Medium

You are given an array nums consisting of positive integers. Split the array into one or more disjoint subarrays such that:

Each element of the array belongs to exactly one subarray
The GCD of the elements of each subarray is strictly greater than 1

Return the minimum number of subarrays that can be obtained after the split.

Note that:

The GCD of a subarray is the largest positive integer that evenly divides all the elements of the subarray.
A subarray is a contiguous part of the array.

Input & Output

Example 1 — Basic Split

$ Input: nums = [12,6,3,14,8]

› Output: 2

💡 Note: Split into [12,6,3] with GCD=3 and [14,8] with GCD=2. Both have GCD > 1, so minimum splits = 2.

Example 2 — Single Element

$ Input: nums = [4]

› Output: 1

💡 Note: Single element [4] has GCD=4 > 1, so we need only 1 subarray.

Example 3 — Each Element Separate

$ Input: nums = [6,10,15]

› Output: 3

💡 Note: GCD(6,10)=2, but adding 15 makes GCD(6,10,15)=1. So split as [6],[10],[15] with 3 subarrays.

Constraints

1 ≤ nums.length ≤ 1000
1 ≤ nums[i] ≤ 10⁹

Visualization

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Asked in

G Google 15 M Microsoft 12 A Amazon 8

The key insight is to use a greedy approach to extend each subarray as long as the GCD remains greater than 1. The optimal approach uses dynamic programming to guarantee the minimum number of splits. Time: O(n³ log m), Space: O(n)

Common Approaches

✓ Brute Force - Try All Splits

⏱️ Time: O(2^n) Space: O(n)

Try every possible way to partition the array into subarrays, check if each partition has all subarrays with GCD > 1, and return the minimum number of subarrays among valid partitions.

Dynamic Programming

⏱️ Time: O(n³ log m) Space: O(n)

Build solution using dynamic programming where dp[i] represents minimum splits needed for array ending at position i. For each position, try all possible starting positions of the last subarray.

Greedy Approach

⏱️ Time: O(n² log m) Space: O(1)

Use a greedy strategy to build subarrays by extending each subarray as long as the GCD remains greater than 1. Start a new subarray when extending would make GCD equal to 1.

Brute Force - Try All Splits — Algorithm Steps

Generate all possible partitions of the array
For each partition, check if all subarrays have GCD > 1
Return minimum number of subarrays among valid partitions

Visualization

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Step-by-Step Walkthrough

Generate Partitions

Create all possible ways to split array

Check GCD

Verify each subarray has GCD > 1

Find Minimum

Return smallest valid partition count

Code -

solution.c — C

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

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

int gcdArray(int* arr, int len) {
    int result = arr[0];
    for (int i = 1; i < len; i++) {
        result = gcd(result, arr[i]);
    }
    return result;
}

int backtrack(int* nums, int n, int start, int currentSplits) {
    if (start == n) {
        return currentSplits;
    }
    
    int minSplits = INT_MAX;
    for (int end = start + 1; end <= n; end++) {
        if (gcdArray(nums + start, end - start) > 1) {
            int result = backtrack(nums, n, end, currentSplits + 1);
            if (result < minSplits) {
                minSplits = result;
            }
        }
    }
    
    return minSplits;
}

int solution(int* nums, int n) {
    if (n == 1) {
        return nums[0] > 1 ? 1 : INT_MAX;
    }
    
    int result = backtrack(nums, n, 0, 0);
    return result == INT_MAX ? -1 : result;
}

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);
    
    int nums[100];
    int n = 0;
    char* token = strtok(line, "[],\n");
    while (token != NULL) {
        nums[n++] = atoi(token);
        token = strtok(NULL, "[],\n");
    }
    
    int result = solution(nums, n);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(2^n)

There are 2^(n-1) ways to partition an array of length n

⚠ Quadratic Growth

Space Complexity

O(n)

Recursion stack depth and temporary arrays for partitions

✓ Linear Space

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Medium Frequency

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Try All Splits — Algorithm Steps

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Code -

Time & Space Complexity

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