Split the Array - Practice Coding Problems

Split the Array - Problem

You're given an integer array nums of even length, and your task is to determine if you can split it into two equal-sized subarrays where each subarray contains only distinct elements.

More specifically, you need to split the array into:

nums1 and nums2 where both have length nums.length / 2
Each element in nums1 must be unique (no duplicates)
Each element in nums2 must be unique (no duplicates)

Return true if such a split is possible, false otherwise.

Example: If nums = [1,1,2,2], you can split it as nums1 = [1,2] and nums2 = [1,2]. Both subarrays have distinct elements, so return true.

Input & Output

example_1.py — Basic Valid Split

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

› Output: true

💡 Note: We can split into nums1 = [1,2] and nums2 = [1,2]. Both arrays have length 2 and contain only distinct elements.

example_2.py — Impossible Split

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

› Output: false

💡 Note: No matter how we split, each array would need 2 elements, but we only have element '1' appearing 4 times. Each sub-array would have duplicates.

example_3.py — Mixed Elements

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

› Output: true

💡 Note: We can split into nums1 = [1,2] and nums2 = [1,3]. Both arrays have distinct elements and equal length.

Constraints

2 ≤ nums.length ≤ 100
nums.length is even
1 ≤ nums[i] ≤ 100

Visualization

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Understanding the Visualization

Count Jersey Numbers

Count how many of each jersey number we have

Check Distribution

Verify if we can give each team unique jersey numbers

Key Takeaway

🎯 Key Insight: Mathematical validation is much faster than trying all combinations - just count frequencies and check if distribution is possible!

Asked in

G Google 15 a Amazon 12 ⊞ Microsoft 8 f Meta 6

The optimal solution uses frequency counting to solve this problem in O(n) time. Key insight: if any element appears more than twice, splitting is impossible. Otherwise, we count elements appearing once vs twice and check if we can distribute them to form two arrays of size n/2 with distinct elements each.

Common Approaches

Approach	Time	Space	Notes
✓ One-Pass Hash Table (Optimal)	O(n)	O(n)	Count frequencies in one pass and check if split is mathematically possible
Brute Force (Generate All Combinations)	O(2^n * n)	O(n)	Generate all possible ways to split the array and check each split

One-Pass Hash Table (Optimal) — Algorithm Steps

Count frequency of each element using hash table
Check if any element appears more than twice (return false if so)
Count how many elements appear exactly once vs twice
Verify if we can form two arrays of size n/2 with distinct elements

Visualization

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Step-by-Step Walkthrough

Count Frequencies

Build hash table with element frequencies

Validate Constraints

Check if split is mathematically possible

Code -

solution.c — C

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

#define MAX_VAL 1001

bool splitArraySameAverage(int* nums, int numsSize) {
    int freq[MAX_VAL * 2] = {0}; // Handle negative numbers by offsetting
    
    // Count frequencies
    for (int i = 0; i < numsSize; i++) {
        freq[nums[i] + MAX_VAL]++;
    }
    
    // Check if any element appears more than twice
    for (int i = 0; i < MAX_VAL * 2; i++) {
        if (freq[i] > 2) {
            return false;
        }
    }
    
    // Count elements that appear once vs twice
    int ones = 0, twos = 0;
    for (int i = 0; i < MAX_VAL * 2; i++) {
        if (freq[i] == 1) ones++;
        else if (freq[i] == 2) twos++;
    }
    
    int target = numsSize / 2;
    return twos <= target && target <= twos + ones;
}

int main() {
    // Parse input and call solution
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through array to count frequencies, then O(1) check

✓ Linear Growth

Space Complexity

O(n)

Hash table to store frequency counts

⚡ Linearithmic Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

One-Pass Hash Table (Optimal) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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