Count Beautiful Splits in an Array - Practice Coding Problems

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 = [1,2,3,1,2]

› Output: 1

💡 Note: Only one valid split: nums1=[1,2], nums2=[3], nums3=[1,2] - nums2 is not prefix of nums3, but we need to check if nums1 is prefix of nums2 (no) or nums2 is prefix of nums3 (no). Actually nums1=[1], nums2=[2,3,1], nums3=[2] gives 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

Tap to expand

Understanding the Visualization

Input Array

Given array nums to be split into 3 non-empty parts

Find Splits

Try all possible ways to create nums1, nums2, nums3

Check Conditions

Count splits where nums1⊆nums2 OR nums2⊆nums3

Key Takeaway

🎯 Key Insight: A split is beautiful when one subarray is a prefix of the next subarray

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

Approach	Time	Space	Notes
✓ Optimized with Preprocessing	O(n³)	O(n²)	Pre-calculate all prefix relationships to avoid repeated comparisons
Brute Force Triple Loop	O(n³)	O(1)	Try all possible splits and check prefix conditions directly

Optimized with Preprocessing — Algorithm Steps

Step 1: Pre-calculate all possible prefix matches between positions
Step 2: Use nested loops for split points but lookup prefix relationships in O(1)
Step 3: Count valid splits using pre-computed results

Visualization

Tap to expand

Step-by-Step Walkthrough

Organized Loops

Systematic iteration through split points

Length Check

Quick length validation before element comparison

Smart Compare

Efficient prefix validation with early termination

Code -

solution.c — C

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

int solution(int* nums, int numsSize) {
    if (numsSize < 3) return 0;
    
    int count = 0;
    
    for (int i = 1; i < numsSize - 1; i++) {
        for (int j = i + 1; j < numsSize; j++) {
            int nums1Len = i;
            int nums2Len = j - i;
            int nums3Len = numsSize - j;
            
            // Check condition 1
            bool condition1 = false;
            if (nums1Len <= nums2Len) {
                condition1 = true;
                for (int k = 0; k < nums1Len; k++) {
                    if (nums[k] != nums[i + k]) {
                        condition1 = false;
                        break;
                    }
                }
            }
            
            // Check condition 2
            bool condition2 = false;
            if (nums2Len <= nums3Len) {
                condition2 = true;
                for (int k = 0; k < nums2Len; k++) {
                    if (nums[i + k] != nums[j + k]) {
                        condition2 = false;
                        break;
                    }
                }
            }
            
            if (condition1 || condition2) {
                count++;
            }
        }
    }
    
    return count;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    int nums[1000];
    int size = 0;
    char* ptr = strchr(line, '[') + 1;
    char* token = strtok(ptr, ",]");
    while (token != NULL) {
        nums[size++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    int result = solution(nums, size);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n³)

Pre-processing is O(n³), main loop is O(n²), dominated by preprocessing

⚠ Quadratic Growth

Space Complexity

O(n²)

Store prefix relationship matrix for all position pairs

⚠ Quadratic Space

28.0K Views

Medium Frequency

~25 min Avg. Time

890 Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Optimized with Preprocessing — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler