LeetCode #3132 — MEDIUM

Find the Integer Added to Array II

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given two integer arrays nums1 and nums2.

From nums1 two elements have been removed, and all other elements have 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 minimum possible integer x that achieves this equivalence.

Example 1:

Input: nums1 = [4,20,16,12,8], nums2 = [14,18,10]

Output: -2

Explanation:

After removing elements at indices [0,4] and adding -2, nums1 becomes [18,14,10].

Example 2:

Input: nums1 = [3,5,5,3], nums2 = [7,7]

Output: 2

Explanation:

After removing elements at indices [0,3] and adding 2, nums1 becomes [7,7].

Constraints:

3 <= nums1.length <= 200
nums2.length == nums1.length - 2
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 removing two elements and adding x to each element of nums1.

Step 01

Brute Force Baseline

Problem summary: You are given two integer arrays nums1 and nums2. From nums1 two elements have been removed, and all other elements have 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 minimum possible integer x that achieves this equivalence.

Baseline thinking

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

Pattern signal: Array · Two Pointers

Example 1

[4,20,16,12,8]
[14,18,10]

Example 2

[3,5,5,3]
[7,7]

Step 02

Core Insight

What unlocks the optimal approach

Try all possibilities to remove 2 elements from <code>nums1</code>.
<code>x</code> should be equal to <code>min(nums2) - min(nums1)</code>, check it naively.

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 #3132: Find the Integer Added to Array II
class Solution {
    public int minimumAddedInteger(int[] nums1, int[] nums2) {
        Arrays.sort(nums1);
        Arrays.sort(nums2);
        int ans = 1 << 30;
        for (int i = 0; i < 3; ++i) {
            int x = nums2[0] - nums1[i];
            if (f(nums1, nums2, x)) {
                ans = Math.min(ans, x);
            }
        }
        return ans;
    }

    private boolean f(int[] nums1, int[] nums2, int x) {
        int i = 0, j = 0, cnt = 0;
        while (i < nums1.length && j < nums2.length) {
            if (nums2[j] - nums1[i] != x) {
                ++cnt;
            } else {
                ++j;
            }
            ++i;
        }
        return cnt <= 2;
    }
}

// Accepted solution for LeetCode #3132: Find the Integer Added to Array II
func minimumAddedInteger(nums1 []int, nums2 []int) int {
	sort.Ints(nums1)
	sort.Ints(nums2)
	ans := 1 << 30
	f := func(x int) bool {
		i, j, cnt := 0, 0, 0
		for i < len(nums1) && j < len(nums2) {
			if nums2[j]-nums1[i] != x {
				cnt++
			} else {
				j++
			}
			i++
		}
		return cnt <= 2
	}
	for _, a := range nums1[:3] {
		x := nums2[0] - a
		if f(x) {
			ans = min(ans, x)
		}
	}
	return ans
}

# Accepted solution for LeetCode #3132: Find the Integer Added to Array II
class Solution:
    def minimumAddedInteger(self, nums1: List[int], nums2: List[int]) -> int:
        def f(x: int) -> bool:
            i = j = cnt = 0
            while i < len(nums1) and j < len(nums2):
                if nums2[j] - nums1[i] != x:
                    cnt += 1
                else:
                    j += 1
                i += 1
            return cnt <= 2

        nums1.sort()
        nums2.sort()
        ans = inf
        for i in range(3):
            x = nums2[0] - nums1[i]
            if f(x):
                ans = min(ans, x)
        return ans

// Accepted solution for LeetCode #3132: Find the Integer Added to Array II
// 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 #3132: Find the Integer Added to Array II
// class Solution {
//     public int minimumAddedInteger(int[] nums1, int[] nums2) {
//         Arrays.sort(nums1);
//         Arrays.sort(nums2);
//         int ans = 1 << 30;
//         for (int i = 0; i < 3; ++i) {
//             int x = nums2[0] - nums1[i];
//             if (f(nums1, nums2, x)) {
//                 ans = Math.min(ans, x);
//             }
//         }
//         return ans;
//     }
// 
//     private boolean f(int[] nums1, int[] nums2, int x) {
//         int i = 0, j = 0, cnt = 0;
//         while (i < nums1.length && j < nums2.length) {
//             if (nums2[j] - nums1[i] != x) {
//                 ++cnt;
//             } else {
//                 ++j;
//             }
//             ++i;
//         }
//         return cnt <= 2;
//     }
// }

// Accepted solution for LeetCode #3132: Find the Integer Added to Array II
function minimumAddedInteger(nums1: number[], nums2: number[]): number {
    nums1.sort((a, b) => a - b);
    nums2.sort((a, b) => a - b);
    const f = (x: number): boolean => {
        let [i, j, cnt] = [0, 0, 0];
        while (i < nums1.length && j < nums2.length) {
            if (nums2[j] - nums1[i] !== x) {
                ++cnt;
            } else {
                ++j;
            }
            ++i;
        }
        return cnt <= 2;
    };
    let ans = Infinity;
    for (let i = 0; i < 3; ++i) {
        const x = nums2[0] - nums1[i];
        if (f(x)) {
            ans = Math.min(ans, x);
        }
    }
    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 n)

Space

O(log n)

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.