Count Beautiful Splits in an Array

Count Beautiful Splits in an Array - Problem

You are given an array nums. A split of an array nums is beautiful if:

The array nums is split into three subarrays: nums1, nums2, and nums3, such that nums can be formed by concatenating nums1, nums2, and nums3 in that order.
The subarray nums1 is a prefix of nums2 OR nums2 is a prefix of nums3.

Return the number of ways you can make this split.

Input & Output

Example 1 — Basic Case

$ Input: nums = [1,1,2,1]

› Output: 2

💡 Note: Split 1: nums1=[1], nums2=[1], nums3=[2,1] - nums1 is prefix of nums2. Split 2: nums1=[1], nums2=[1,2], nums3=[1] - nums1 is prefix of nums2.

Example 2 — Single Valid Split

$ Input: nums = [1,2,1,2,1]

› Output: 2

💡 Note: Split 1: nums1=[1], nums2=[2,1,2], nums3=[1] - nums2 is prefix of nums3. Split 2: nums1=[1,2,1], nums2=[2], nums3=[1] - nums2 is prefix of nums3.

Example 3 — No Valid Splits

$ Input: nums = [1,2,3]

› Output: 0

💡 Note: Only possible split: nums1=[1], nums2=[2], nums3=[3] - neither nums1 is prefix of nums2 nor nums2 is prefix of nums3.

Constraints

3 ≤ nums.length ≤ 1000
1 ≤ nums[i] ≤ 40

Visualization

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Asked in

G Google 15 a Amazon 12

The key insight is to systematically check all possible 3-way splits and verify if nums1 is a prefix of nums2 OR nums2 is a prefix of nums3. Best approach uses nested loops with efficient prefix checking. Time: O(n³), Space: O(1)

Common Approaches

✓ Brute Force Triple Loop

⏱️ Time: O(n³) Space: O(1)

Generate all possible ways to split the array into 3 non-empty parts, then for each split check if nums1 is a prefix of nums2 OR nums2 is a prefix of nums3 by comparing elements one by one.

Optimized with Preprocessing

⏱️ Time: O(n³) Space: O(n²)

Instead of checking prefix relationships repeatedly, pre-compute for each position whether it can be the start of a matching prefix. This reduces redundant comparisons in the inner loops.

Brute Force Triple Loop — Algorithm Steps

Step 1: Try all possible split points i and j where i < j
Step 2: For each split, extract nums1[0:i], nums2[i:j], nums3[j:n]
Step 3: Check if nums1 matches start of nums2 OR nums2 matches start of nums3

Visualization

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Step-by-Step Walkthrough

Split Points

Try all i,j where 1≤i<j<n

Extract Parts

Get nums1[0:i], nums2[i:j], nums3[j:n]

Check Prefix

Verify nums1⊆nums2 OR nums2⊆nums3

Code -

solution.c — C

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

int solution(int* nums, int numsSize) {
    int count = 0;
    
    for (int i = 1; i < numsSize - 1; i++) {
        for (int j = i + 1; j < numsSize; j++) {
            // Check if nums1 is prefix of nums2
            bool isPrefix12 = false;
            if (i <= j - i) {
                isPrefix12 = true;
                for (int k = 0; k < i; k++) {
                    if (nums[k] != nums[i + k]) {
                        isPrefix12 = false;
                        break;
                    }
                }
            }
            
            // Check if nums2 is prefix of nums3
            bool isPrefix23 = false;
            int nums2Len = j - i;
            int nums3Len = numsSize - j;
            if (nums2Len <= nums3Len) {
                isPrefix23 = true;
                for (int k = 0; k < nums2Len; k++) {
                    if (nums[i + k] != nums[j + k]) {
                        isPrefix23 = false;
                        break;
                    }
                }
            }
            
            if (isPrefix12 || isPrefix23) {
                count++;
            }
        }
    }
    
    return count;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    // Parse array using strtol
    int nums[1000];
    int size = 0;
    char* ptr = line;
    
    // Skip to first digit
    while (*ptr && (*ptr < '0' || *ptr > '9')) ptr++;
    
    // Parse numbers
    while (*ptr && size < 1000) {
        char* endptr;
        nums[size++] = (int)strtol(ptr, &endptr, 10);
        ptr = endptr;
        // Skip to next digit
        while (*ptr && (*ptr < '0' || *ptr > '9')) ptr++;
        if (*ptr == '\0' || *ptr == ']') break;
    }
    
    int result = solution(nums, size);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n³)

Nested loops O(n²) for split points, plus O(n) for prefix comparison

⚠ Quadratic Growth

Space Complexity

O(1)

Only using constant extra variables for indices and comparisons

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force Triple Loop — Algorithm Steps

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Code -

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