Find the Array Concatenation Value

Find the Array Concatenation Value - Problem

You are given a 0-indexed integer array nums.

The concatenation of two numbers is the number formed by concatenating their numerals.

For example, the concatenation of 15, 49 is 1549.

The concatenation value of nums is initially equal to 0. Perform this operation until nums becomes empty:

If nums has a size greater than one, add the value of the concatenation of the first and the last element to the concatenation value of nums, and remove those two elements from nums.
If only one element exists in nums, add its value to the concatenation value of nums, then remove it.

Return the concatenation value of nums.

Input & Output

Example 1 — Basic Case

$ Input: nums = [7,52,2,4]

› Output: 596

💡 Note: First iteration: concatenate 7 and 4 to get 74. Second iteration: concatenate 52 and 2 to get 522. Total: 74 + 522 = 596

Example 2 — Odd Length Array

$ Input: nums = [5,14,13,8,12]

› Output: 673

💡 Note: Pair 5+12=512, pair 14+8=148, middle element 13 remains. Total: 512 + 148 + 13 = 673

Example 3 — Single Element

$ Input: nums = [15]

› Output: 15

💡 Note: Only one element exists, so return it as is: 15

Constraints

1 ≤ nums.length ≤ 1000
1 ≤ nums[i] ≤ 10⁴

Visualization

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Asked in

G Google 15 a Amazon 12 M Microsoft 8

The key insight is to use two pointers to access elements from both ends without modifying the array. Best approach is the two-pointers method with string concatenation. Time: O(n), Space: O(1)

Common Approaches

✓ Brute Force with Array Modification

⏱️ Time: O(n²) Space: O(1)

Modify the array by removing elements from both ends, convert numbers to strings for concatenation, then convert back to integers and add to result.

Two Pointers Approach

⏱️ Time: O(n) Space: O(1)

Use left and right pointers to access elements from both ends of the array without modifying it. Move pointers inward after each concatenation.

Brute Force with Array Modification — Algorithm Steps

While array has more than 1 element, remove first and last
Convert both to strings and concatenate
Convert result back to integer and add to sum
If one element remains, add it to sum

Visualization

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Step-by-Step Walkthrough

Initial Array

Start with array [7,52,2,4]

Remove & Concatenate

Remove 7 and 4, concatenate to get 74

Continue Process

Remove remaining elements and sum concatenations

Code -

solution.c — C

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

long long solution(int* nums, int size) {
    long long result = 0;
    int* temp = malloc(size * sizeof(int));
    memcpy(temp, nums, size * sizeof(int));
    
    int left = 0, right = size - 1;
    
    while (left < right) {
        char concatenated[20];
        sprintf(concatenated, "%d%d", temp[left], temp[right]);
        result += atoll(concatenated);
        left++;
        right--;
    }
    
    if (left == right) {
        result += temp[left];
    }
    
    free(temp);
    return result;
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    int nums[100];
    int size = 0;
    char* token = strtok(line + 1, ",]");
    while (token != NULL) {
        nums[size++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    printf("%lld\n", solution(nums, size));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Each removal operation takes O(n) time and we do this n/2 times

⚠ Quadratic Growth

Space Complexity

O(1)

Only using a few variables for concatenation

✓ Linear Space

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Medium Frequency

~15 min Avg. Time

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Smart Actions

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force with Array Modification — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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