LeetCode #3131 — EASY

Find the Integer Added to Array I

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

The Problem

Problem Statement

You are given two arrays of equal length, nums1 and nums2.

Each element in nums1 has been increased (or decreased in the case of negative) by an integer, represented by the variable x.

As a result, nums1 becomes equal to nums2. Two arrays are considered equal when they contain the same integers with the same frequencies.

Return the integer x.

Example 1:

Input: nums1 = [2,6,4], nums2 = [9,7,5]

Output: 3

Explanation:

The integer added to each element of nums1 is 3.

Example 2:

Input: nums1 = [10], nums2 = [5]

Output: -5

Explanation:

The integer added to each element of nums1 is -5.

Example 3:

Input: nums1 = [1,1,1,1], nums2 = [1,1,1,1]

Output: 0

Explanation:

The integer added to each element of nums1 is 0.

Constraints:

1 <= nums1.length == nums2.length <= 100
0 <= nums1[i], nums2[i] <= 1000
The test cases are generated in a way that there is an integer x such that nums1 can become equal to nums2 by adding x to each element of nums1.

Step 01

Brute Force Baseline

Problem summary: You are given two arrays of equal length, nums1 and nums2. Each element in nums1 has been increased (or decreased in the case of negative) by an integer, represented by the variable x. As a result, nums1 becomes equal to nums2. Two arrays are considered equal when they contain the same integers with the same frequencies. Return the integer x.

Baseline thinking

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

Pattern signal: Array

Example 1

[2,6,4]
[9,7,5]

Example 2

[10]
[5]

Example 3

[1,1,1,1]
[1,1,1,1]

Step 02

Core Insight

What unlocks the optimal approach

Notice that, after sorting both arrays, there should be a one-to-one correspondence between every element.
Thus <code>x = min(nums2) - min(nums1)</code>.

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 #3131: Find the Integer Added to Array I
class Solution {
    public int addedInteger(int[] nums1, int[] nums2) {
        return Arrays.stream(nums2).min().getAsInt() - Arrays.stream(nums1).min().getAsInt();
    }
}

// Accepted solution for LeetCode #3131: Find the Integer Added to Array I
func addedInteger(nums1 []int, nums2 []int) int {
	return slices.Min(nums2) - slices.Min(nums1)
}

# Accepted solution for LeetCode #3131: Find the Integer Added to Array I
class Solution:
    def addedInteger(self, nums1: List[int], nums2: List[int]) -> int:
        return min(nums2) - min(nums1)

// Accepted solution for LeetCode #3131: Find the Integer Added to Array I
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3131: Find the Integer Added to Array I
// class Solution {
//     public int addedInteger(int[] nums1, int[] nums2) {
//         return Arrays.stream(nums2).min().getAsInt() - Arrays.stream(nums1).min().getAsInt();
//     }
// }

// Accepted solution for LeetCode #3131: Find the Integer Added to Array I
function addedInteger(nums1: number[], nums2: number[]): number {
    return Math.min(...nums2) - Math.min(...nums1);
}

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)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

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.