Even-Odd Separator - Problem

Given an array of integers, separate the even and odd numbers into two different arrays while maintaining their original order.

You need to return a 2D array where:

The first sub-array contains all even numbers in their original order
The second sub-array contains all odd numbers in their original order

Example: For input [1, 2, 3, 4, 5, 6], return [[2, 4, 6], [1, 3, 5]]

Input & Output

Example 1 — Basic Case

$ Input: nums = [1,2,3,4,5,6]

› Output: [[2,4,6],[1,3,5]]

💡 Note: Even numbers [2,4,6] maintain original order, odd numbers [1,3,5] maintain original order

Example 2 — All Even Numbers

$ Input: nums = [2,4,6,8]

› Output: [[2,4,6,8],[]]

💡 Note: All numbers are even, so first array contains all elements, second array is empty

Example 3 — Mixed with Negatives

$ Input: nums = [-3,-2,-1,0,1,2]

› Output: [[-2,0,2],[-3,-1,1]]

💡 Note: Even numbers include negatives and zero: [-2,0,2], odds: [-3,-1,1]

Constraints

1 ≤ nums.length ≤ 10³
-10⁶ ≤ nums[i] ≤ 10⁶

Visualization

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Asked in

a Amazon 25 M Microsoft 18 G Google 15 A Apple 12

The key insight is to use a single pass through the array to separate even and odd numbers simultaneously. The optimal approach iterates once, checking each number's parity and adding it to the appropriate result array. Time: O(n), Space: O(n)

Common Approaches

✓ Single Pass Separation

⏱️ Time: O(n) Space: O(n)

Use a single loop to traverse the array, checking each number's parity and adding it to the appropriate result array in one pass.

Two Separate Loops

⏱️ Time: O(n) Space: O(n)

First loop finds all even numbers and adds them to the result array, then second loop finds all odd numbers and adds them to the result array.

Single Pass Separation — Algorithm Steps

Initialize two empty arrays for even and odd numbers
Iterate through input array once
For each number, check if even or odd and add to appropriate array
Return both arrays as 2D result

Visualization

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Step-by-Step Walkthrough

Initialize

Create two empty arrays for results

Single Pass

Check each element and add to appropriate array

Result

Both arrays populated in one iteration

Code -

solution.c — C

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

void solution(int* nums, int numsSize, int** result, int* returnSize, int** returnColumnSizes) {
    int* evens = (int*)malloc(numsSize * sizeof(int));
    int* odds = (int*)malloc(numsSize * sizeof(int));
    int evenCount = 0, oddCount = 0;
    
    // Single pass: separate even and odd numbers
    for (int i = 0; i < numsSize; i++) {
        if (nums[i] % 2 == 0) {
            evens[evenCount++] = nums[i];
        } else {
            odds[oddCount++] = nums[i];
        }
    }
    
    result[0] = evens;
    result[1] = odds;
    (*returnColumnSizes)[0] = evenCount;
    (*returnColumnSizes)[1] = oddCount;
    *returnSize = 2;
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    int nums[100], numsSize = 0;
    char* token = strtok(line + 1, ",]");
    while (token != NULL) {
        nums[numsSize++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    int** result = (int**)malloc(2 * sizeof(int*));
    int* returnColumnSizes = (int*)malloc(2 * sizeof(int));
    int returnSize;
    
    solution(nums, numsSize, result, &returnSize, &returnColumnSizes);
    
    printf("[");
    for (int i = 0; i < returnSize; i++) {
        printf("[");
        for (int j = 0; j < returnColumnSizes[i]; j++) {
            printf("%d", result[i][j]);
            if (j < returnColumnSizes[i] - 1) printf(",");
        }
        printf("]");
        if (i < returnSize - 1) printf(",");
    }
    printf("]\n");
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single loop through the array, each element processed once

✓ Linear Growth

Space Complexity

O(n)

Two result arrays to store even and odd numbers, total space proportional to input size

⚡ Linearithmic Space

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Medium Frequency

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Even-Odd Separator - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Single Pass Separation — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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