LeetCode #2562 — EASY

Find the Array Concatenation Value

Build confidence with an intuition-first walkthrough focused on array fundamentals.

Solve on LeetCode

The Problem

Problem Statement

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. For example, if the nums was [1, 2, 4, 5, 6], add 16 to the concatenation value.
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.

Example 1:

Input: nums = [7,52,2,4]
Output: 596
Explanation: Before performing any operation, nums is [7,52,2,4] and concatenation value is 0.
 - In the first operation:
We pick the first element, 7, and the last element, 4.
Their concatenation is 74, and we add it to the concatenation value, so it becomes equal to 74.
Then we delete them from nums, so nums becomes equal to [52,2].
 - In the second operation:
We pick the first element, 52, and the last element, 2.
Their concatenation is 522, and we add it to the concatenation value, so it becomes equal to 596.
Then we delete them from the nums, so nums becomes empty.
Since the concatenation value is 596 so the answer is 596.

Example 2:

Input: nums = [5,14,13,8,12]
Output: 673
Explanation: Before performing any operation, nums is [5,14,13,8,12] and concatenation value is 0.
 - In the first operation:
We pick the first element, 5, and the last element, 12.
Their concatenation is 512, and we add it to the concatenation value, so it becomes equal to 512.
Then we delete them from the nums, so nums becomes equal to [14,13,8].
 - In the second operation:
We pick the first element, 14, and the last element, 8.
Their concatenation is 148, and we add it to the concatenation value, so it becomes equal to 660.
Then we delete them from the nums, so nums becomes equal to [13].
 - In the third operation:
nums has only one element, so we pick 13 and add it to the concatenation value, so it becomes equal to 673.
Then we delete it from nums, so nums become empty.
Since the concatenation value is 673 so the answer is 673.

Constraints:

1 <= nums.length <= 1000
1 <= nums[i] <= 10⁴

Step 01

Brute Force Baseline

Problem summary: 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. For example, if the nums was [1, 2, 4, 5, 6], add 16 to the concatenation value. 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Two Pointers

Example 1

[7,52,2,4]

Example 2

[5,14,13,8,12]

Step 02

Core Insight

What unlocks the optimal approach

Consider simulating the process to calculate the answer
iterate until the array becomes empty. In each iteration, concatenate the first element to the last element and add their concatenation value to the answer.
Don’t forget to handle cases when one element is left in the end, not two elements.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2562: Find the Array Concatenation Value
class Solution {
    public long findTheArrayConcVal(int[] nums) {
        long ans = 0;
        int i = 0, j = nums.length - 1;
        for (; i < j; ++i, --j) {
            ans += Integer.parseInt(nums[i] + "" + nums[j]);
        }
        if (i == j) {
            ans += nums[i];
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2562: Find the Array Concatenation Value
func findTheArrayConcVal(nums []int) (ans int64) {
	i, j := 0, len(nums)-1
	for ; i < j; i, j = i+1, j-1 {
		x, _ := strconv.Atoi(strconv.Itoa(nums[i]) + strconv.Itoa(nums[j]))
		ans += int64(x)
	}
	if i == j {
		ans += int64(nums[i])
	}
	return
}

# Accepted solution for LeetCode #2562: Find the Array Concatenation Value
class Solution:
    def findTheArrayConcVal(self, nums: List[int]) -> int:
        ans = 0
        i, j = 0, len(nums) - 1
        while i < j:
            ans += int(str(nums[i]) + str(nums[j]))
            i, j = i + 1, j - 1
        if i == j:
            ans += nums[i]
        return ans

// Accepted solution for LeetCode #2562: Find the Array Concatenation Value
impl Solution {
    pub fn find_the_array_conc_val(nums: Vec<i32>) -> i64 {
        let n = nums.len();
        let mut ans = 0;
        let mut i = 0;
        let mut j = n - 1;
        while i < j {
            ans += format!("{}{}", nums[i], nums[j]).parse::<i64>().unwrap();
            i += 1;
            j -= 1;
        }
        if i == j {
            ans += nums[i] as i64;
        }
        ans
    }
}

// Accepted solution for LeetCode #2562: Find the Array Concatenation Value
function findTheArrayConcVal(nums: number[]): number {
    const n = nums.length;
    let ans = 0;
    let i = 0;
    let j = n - 1;
    while (i < j) {
        ans += Number(`${nums[i]}${nums[j]}`);
        i++;
        j--;
    }
    if (i === j) {
        ans += nums[i];
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × log M)

Space

O(log M)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.

TWO POINTERS

O(n) time

O(1) space

Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.

Shortcut: Two converging pointers on sorted data → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Moving both pointers on every comparison

Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.

Usually fails on: A valid pair can be skipped when only one side should move.

Fix: Move exactly one pointer per decision branch based on invariant.