Concatenation of Array

Concatenation of Array - Problem

Array Simulation Easy

Given an integer array nums of length n, you want to create an array ans of length 2n where ans[i] == nums[i] and ans[i + n] == nums[i] for 0 <= i < n (0-indexed).

Specifically, ans is the concatenation of two nums arrays.

Return the array ans.

Input & Output

Example 1 — Basic Case

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

› Output: [1,2,1,1,2,1]

💡 Note: The array ans is [1,2,1,1,2,1]. ans = [nums[0],nums[1],nums[2],nums[0],nums[1],nums[2]] = [1,2,1,1,2,1]

Example 2 — Single Element

$ Input: nums = [1]

› Output: [1,1]

💡 Note: The array ans is [1,1]. ans = [nums[0],nums[0]] = [1,1]

Example 3 — Multiple Elements

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

› Output: [1,3,2,1,1,3,2,1]

💡 Note: The array ans is [1,3,2,1,1,3,2,1]. The original array is concatenated with itself

Constraints

n == nums.length
1 ≤ n ≤ 1000
1 ≤ nums[i] ≤ 1000

Visualization

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Asked in

a Amazon 12 🍎 Apple 8

The key insight is to create a result array of size 2n and copy each element from the original array to two positions: index i and index i+n. The optimal single-pass approach handles both copies in one loop. Time: O(n), Space: O(n).

Common Approaches

✓ Brute Force - Element by Element

⏱️ Time: O(n) Space: O(n)

First loop through the array to add all elements once, then loop through again to add all elements a second time. This approach uses two separate passes through the original array.

Single Pass Concatenation

⏱️ Time: O(n) Space: O(n)

In a single loop through the original array, copy each element to both its first position (index i) and second position (index i+n) in the result array. This is more efficient than two separate passes.

Brute Force - Element by Element — Algorithm Steps

Create result array of size 2n
First pass: copy nums[i] to ans[i]
Second pass: copy nums[i] to ans[i+n]

Visualization

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Step-by-Step Walkthrough

First Pass

Copy original elements to positions 0 to n-1

Second Pass

Copy original elements again to positions n to 2n-1

Result

Array contains nums concatenated with itself

Code -

solution.c — C

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

int* solution(int* nums, int numsSize, int* returnSize) {
    *returnSize = 2 * numsSize;
    int* ans = (int*)malloc(*returnSize * sizeof(int));
    
    // First pass: copy original elements
    for (int i = 0; i < numsSize; i++) {
        ans[i] = nums[i];
    }
    
    // Second pass: copy elements again at offset n
    for (int i = 0; i < numsSize; i++) {
        ans[i + numsSize] = nums[i];
    }
    
    return ans;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    // Parse array manually
    int nums[1000];
    int numsSize = 0;
    char* ptr = line;
    while (*ptr) {
        if (*ptr >= '0' && *ptr <= '9' || *ptr == '-') {
            nums[numsSize++] = atoi(ptr);
            while (*ptr && *ptr != ',' && *ptr != ']') ptr++;
        } else {
            ptr++;
        }
    }
    
    int returnSize;
    int* result = solution(nums, numsSize, &returnSize);
    
    printf("[");
    for (int i = 0; i < returnSize; i++) {
        printf("%d", result[i]);
        if (i < returnSize - 1) printf(",");
    }
    printf("]\n");
    
    free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Two loops each iterating n times: O(n) + O(n) = O(n)

✓ Linear Growth

Space Complexity

O(n)

Result array stores 2n elements, which is O(n) space

⚡ Linearithmic Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Element by Element — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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