Sort Array By Parity II

Sort Array By Parity II - Problem

Given an array of integers nums, half of the integers in nums are odd, and the other half are even.

Sort the array so that whenever nums[i] is odd, i is odd, and whenever nums[i] is even, i is even.

Return any answer array that satisfies this condition.

Input & Output

Example 1 — Basic Case

$ Input: nums = [4,2,5,7]

› Output: [4,5,2,7]

💡 Note: Index 0 (even) has 4 (even) ✓, index 1 (odd) needs odd number, so we place 5. Index 2 (even) gets remaining even 2, index 3 (odd) gets remaining odd 7.

Example 2 — Already Correct

$ Input: nums = [2,3]

› Output: [2,3]

💡 Note: Index 0 (even) has 2 (even) ✓, index 1 (odd) has 3 (odd) ✓. Array is already correctly arranged.

Example 3 — Larger Array

$ Input: nums = [4,2,5,7,8,9]

› Output: [4,5,2,7,8,9]

💡 Note: Even indices (0,2,4) should have even numbers (4,2,8) and odd indices (1,3,5) should have odd numbers (5,7,9). One valid arrangement is [4,5,2,7,8,9].

Constraints

2 ≤ nums.length ≤ 2 × 10⁴
nums.length is even
Half of integers in nums are even
0 ≤ nums[i] ≤ 1000

Visualization

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Asked in

f Facebook 15 🍎 Apple 12 G Google 8 a Amazon 6

The key insight is that even numbers must be at even indices and odd numbers at odd indices. The optimal two-pointers approach swaps misplaced elements in-place. Time: O(n), Space: O(1).

Common Approaches

✓ Separate and Merge

⏱️ Time: O(n) Space: O(n)

First pass collects all even numbers and all odd numbers separately. Second pass places even numbers at even indices (0,2,4...) and odd numbers at odd indices (1,3,5...).

Two Pointers In-Place

⏱️ Time: O(n) Space: O(1)

Maintain two pointers - one for even indices and one for odd indices. When we find a misplaced element, find its correct partner and swap them. Continue until both pointers have processed all positions.

Separate and Merge — Algorithm Steps

Iterate through array and separate odds and evens
Place evens at indices 0,2,4,... and odds at indices 1,3,5,...

Visualization

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Step-by-Step Walkthrough

Separate

Split into even and odd arrays

Place Evens

Put evens at indices 0,2,4,...

Place Odds

Put odds at indices 1,3,5,...

Code -

solution.c — C

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

void solution(int* nums, int numsSize, int* result, int* isValid) {
    static int evens[1000], odds[1000];
    int evenCount = 0, oddCount = 0;
    
    // Separate odd and even numbers
    for (int i = 0; i < numsSize; i++) {
        if (nums[i] % 2 == 0) {
            evens[evenCount++] = nums[i];
        } else {
            odds[oddCount++] = nums[i];
        }
    }
    
    // Check if we have equal number of evens and odds
    if (evenCount != oddCount || evenCount != numsSize / 2) {
        *isValid = 0;
        return;
    }
    
    *isValid = 1;
    
    // Place evens at even indices
    int evenIdx = 0;
    for (int i = 0; i < numsSize; i += 2) {
        result[i] = evens[evenIdx++];
    }
    
    // Place odds at odd indices
    int oddIdx = 0;
    for (int i = 1; i < numsSize; i += 2) {
        result[i] = odds[oddIdx++];
    }
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    // Parse array using proper method
    static int nums[1000];
    int numsSize = 0;
    char* p = line;
    
    // Skip to first digit or bracket
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    
    // Parse numbers
    while (*p && *p != ']') {
        while (*p == ' ' || *p == ',') p++;
        if (*p == ']' || *p == '\0') break;
        
        char* endptr;
        long val = strtol(p, &endptr, 10);
        if (endptr != p) {
            nums[numsSize++] = (int)val;
            p = endptr;
        } else {
            p++;
        }
    }
    
    static int result[1000];
    int isValid;
    solution(nums, numsSize, result, &isValid);
    
    if (!isValid) {
        printf("Invalid test case - input has unequal number of even/odd numbers, making valid solution impossible\n");
    } else {
        printf("[");
        for (int i = 0; i < numsSize; i++) {
            if (i > 0) printf(",");
            printf("%d", result[i]);
        }
        printf("]\n");
    }
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass to separate + single pass to merge = O(n)

✓ Linear Growth

Space Complexity

O(n)

Arrays to store separated odd and even numbers

⚡ Linearithmic Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Separate and Merge — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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