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Find the Integer Added to Array I (LeetCode)

Problem Description​

Problem StatementSolution LinkLeetCode Profile
Find the Integer Added to Array IFind the Integer Added to Array I Solution on LeetCodevaishu_1904

Problem Description​

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.

Approach​

To find the integer x that was added to each element of nums1 to obtain nums2, we can use the following approach:

  1. Iterate through the arrays nums1 and nums2.
  2. Calculate the difference x for the first pair of elements from nums1 and nums2.
  3. Ensure that the calculated x is consistent for all elements.

Solution Code​

Python​

def find_integer_added_to_array(nums1, nums2):
    if len(nums1) != len(nums2):
        return None  # Inconsistent input lengths
    
    x = nums2[0] - nums1[0]
    
    for i in range(1, len(nums1)):
        if nums2[i] - nums1[i] != x:
            return None  # Inconsistent difference found
    
    return x

C++​

#include <vector>
#include <iostream>
using namespace std;

class Solution {
public:
    int findIntegerAddedToArray(vector<int>& nums1, vector<int>& nums2) {
        if (nums1.size() != nums2.size()) {
            return INT_MIN;  // Inconsistent input lengths
        }
        
        int x = nums2[0] - nums1[0];
        
        for (size_t i = 1; i < nums1.size(); ++i) {
            if (nums2[i] - nums1[i] != x) {
                return INT_MIN;  // Inconsistent difference found
            }
        }
        
        return x;
    }
};

Java​

class Solution {
    public Integer findIntegerAddedToArray(int[] nums1, int[] nums2) {
        if (nums1.length != nums2.length) {
            return null;  // Inconsistent input lengths
        }
        
        int x = nums2[0] - nums1[0];
        
        for (int i = 1; i < nums1.length; i++) {
            if (nums2[i] - nums1[i] != x) {
                return null;  // Inconsistent difference found
            }
        }
        
        return x;
    }

Conclusion​

The above solutions determine the integer x that was added to each element of nums1 to obtain nums2. By ensuring consistency across the entire array, we validate the difference and return the result. Adjustments for different languages and edge cases ensure robustness across various inputs.