Description:
Given two integer arrays nums1 and nums2, return an array of their intersection. Each element in the result must appear as many times as it shows in both arrays and you may return the result in any order.
Examples:
Example 1:
Input: nums1 = [1,2,2,1], nums2 = [2,2]
Output: [2,2]
Example 2:
Input: nums1 = [4,9,5], nums2 = [9,4,9,8,4]
Output: [4,9]
Explanation: [9,4] is also accepted.Solution in Python:
To solve the problem of finding the intersection of two integer arrays where each element in the result appears as many times as it shows in both arrays, we can use a hashmap (dictionary) to count the occurrences of each element in one of the arrays. Then, we iterate through the other array to determine how many times each element should appear in the result based on the counts in the hashmap.
Python
from typing import List
class Solution:
def intersect(self, nums1: List[int], nums2: List[int]) -> List[int]:
# Dictionary to count occurrences in nums1
counts = {}
# Fill the dictionary with counts of each element in nums1
for num in nums1:
if num in counts:
counts[num] += 1
else:
counts[num] = 1
# List to store the result
result = []
# Check each element in nums2 against the dictionary
for num in nums2:
if num in counts and counts[num] > 0:
result.append(num)
counts[num] -= 1
return result
# Example usage:
# Create an instance of the Solution class
solution = Solution()
# Example 1
nums1 = [1, 2, 2, 1]
nums2 = [2, 2]
print(solution.intersect(nums1, nums2)) # Output: [2, 2]
# Example 2
nums1 = [4, 9, 5]
nums2 = [9, 4, 9, 8, 4]
print(solution.intersect(nums1, nums2)) # Output: [4, 9]Explanation of the Code:
- Line 1: We import
Listfromtypingto specify the type of the input lists. - Line 3-5: We define the
Solutionclass and theintersectmethod which takes two listsnums1andnums2as inputs and returns a list of their intersection. - Line 7-11: We create a dictionary
countsto store the counts of each element innums1. - Line 9-11: We iterate over
nums1and update the dictionary with the count of each element. - Line 14: We initialize an empty list
resultto store the intersection result. - Line 17-21: We iterate over
nums2and for each element, check if it is in the dictionary and its count is greater than zero. If so, we append the element toresultand decrease the count in the dictionary. - Line 23: We return the
resultlist containing the intersection elements with the correct counts.
The provided example usages demonstrate how to use the Solution class and the intersect method to get the intersection of two arrays.
Solution in Javascript:
JavaScript
/**
* @param {number[]} nums1
* @param {number[]} nums2
* @return {number[]}
*/
var intersect = function(nums1, nums2) {
// Object to count occurrences in nums1
const counts = {};
// Fill the object with counts of each element in nums1
for (let num of nums1) {
if (counts[num] !== undefined) {
counts[num]++;
} else {
counts[num] = 1;
}
}
// Array to store the result
const result = [];
// Check each element in nums2 against the object
for (let num of nums2) {
if (counts[num] > 0) {
result.push(num);
counts[num]--;
}
}
return result;
};
// Example usage:
// Example 1
let nums1 = [1, 2, 2, 1];
let nums2 = [2, 2];
console.log(intersect(nums1, nums2)); // Output: [2, 2]
// Example 2
nums1 = [4, 9, 5];
nums2 = [9, 4, 9, 8, 4];
console.log(intersect(nums1, nums2)); // Output: [4, 9]Solution in Java:
Java
import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Arrays;
class Solution {
public int[] intersect(int[] nums1, int[] nums2) {
// HashMap to count occurrences in nums1
HashMap<Integer, Integer> counts = new HashMap<>();
// Fill the HashMap with counts of each element in nums1
for (int num : nums1) {
counts.put(num, counts.getOrDefault(num, 0) + 1);
}
// List to store the result
List<Integer> result = new ArrayList<>();
// Check each element in nums2 against the HashMap
for (int num : nums2) {
if (counts.containsKey(num) && counts.get(num) > 0) {
result.add(num);
counts.put(num, counts.get(num) - 1);
}
}
// Convert the result list to an array
int[] resultArray = new int[result.size()];
for (int i = 0; i < result.size(); i++) {
resultArray[i] = result.get(i);
}
return resultArray;
}
public static void main(String[] args) {
Solution solution = new Solution();
// Example 1
int[] nums1Example1 = {1, 2, 2, 1};
int[] nums2Example1 = {2, 2};
int[] result1 = solution.intersect(nums1Example1, nums2Example1);
System.out.println(Arrays.toString(result1)); // Output: [2, 2]
// Example 2
int[] nums1Example2 = {4, 9, 5};
int[] nums2Example2 = {9, 4, 9, 8, 4};
int[] result2 = solution.intersect(nums1Example2, nums2Example2);
System.out.println(Arrays.toString(result2)); // Output: [4, 9]
}
}Solution in C#:
C#
using System;
using System.Collections.Generic;
public class Solution {
public int[] Intersect(int[] nums1, int[] nums2) {
// Dictionary to count occurrences in nums1
Dictionary<int, int> counts = new Dictionary<int, int>();
// Fill the dictionary with counts of each element in nums1
foreach (int num in nums1) {
if (counts.ContainsKey(num)) {
counts[num]++;
} else {
counts.Add(num, 1);
}
}
// List to store the result
List<int> result = new List<int>();
// Check each element in nums2 against the dictionary
foreach (int num in nums2) {
if (counts.ContainsKey(num) && counts[num] > 0) {
result.Add(num);
counts[num]--;
}
}
// Convert the result list to an array
return result.ToArray();
}
}