从具有相同位数总和的两个数组中计数不同的对
原文:https://www . geesforgeks . org/count-distinct-pairs-from-two-array-具有相同的数字总和/
给定两个数组 arr1[]和 arr2[]。任务是找到个不同的对的总数(从 arr1 中选取 1 个元素，从 arr2 中选取 1 个元素)，这样这一对的两个元素都有数字的总和。 注意:出现一次以上的对必须只统计一次。 示例 :
Input : arr1[] = {33, 41, 59, 1, 3}
arr2[] = {3, 32, 51, 3}
Output : 3
Possible pairs are:
(33, 51), (41, 32), (3, 3)
Input : arr1[] = {1, 6, 4, 22}
arr2[] = {1, 3, 24}
Output : 2
Possible pairs are:
(1, 1), (6, 24)
进场:
- 运行两个嵌套循环，从两个数组中生成所有可能的对，从 arr1[]中获取一个元素，从 arr2[]中获取一个元素。
- 如果数字总和相等，则将对(a，b)插入一个集合，以避免重复，其中 a 是较小的元素，b 是较大的元素。
- 总对将是最终集的大小。
以下是上述方法的实现:
C++
// C++ program to count total number of
// pairs having elements with same
// sum of digits
#include <bits/stdc++.h>
using namespace std;
// Function for returning
// sum of digits of a number
int digitSum(int n)
{
int sum = 0;
while (n > 0) {
sum += n % 10;
n = n / 10;
}
return sum;
}
// Function to return the total pairs
// of elements with equal sum of digits
int totalPairs(int arr1[], int arr2[], int n, int m)
{
// set is used to avoid duplicate pairs
set<pair<int, int> > s;
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
// check sum of digits
// of both the elements
if (digitSum(arr1[i]) == digitSum(arr2[j])) {
if (arr1[i] < arr2[j])
s.insert(make_pair(arr1[i], arr2[j]));
else
s.insert(make_pair(arr2[j], arr1[i]));
}
}
}
// return size of the set
return s.size();
}
// Driver code
int main()
{
int arr1[] = { 100, 3, 7, 50 };
int arr2[] = { 5, 1, 10, 4 };
int n = sizeof(arr1) / sizeof(arr1[0]);
int m = sizeof(arr2) / sizeof(arr2[0]);
cout << totalPairs(arr1, arr2, n, m);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to count total number of
// pairs having elements with same
// sum of digits
import java.util.*;
class GFG
{
static class pair
{
int first, second;
public pair(int first, int second)
{
this.first = first;
this.second = second;
}
}
// Function for returning
// sum of digits of a number
static int digitSum(int n)
{
int sum = 0;
while (n > 0)
{
sum += n % 10;
n = n / 10;
}
return sum;
}
// Function to return the total pairs
// of elements with equal sum of digits
static int totalPairs(int arr1[], int arr2[],
int n, int m)
{
// set is used to avoid duplicate pairs
Set<pair> s = new HashSet<>();
for (int i = 0; i < n; i++)
{
for (int j = 0; j < m; j++)
{
// check sum of digits
// of both the elements
if (digitSum(arr1[i]) == digitSum(arr2[j]))
{
if (arr1[i] < arr2[j])
s.add(new pair(arr1[i], arr2[j]));
else
s.add(new pair(arr2[j], arr1[i]));
}
}
}
// return size of the set
return s.size();
}
// Driver code
public static void main(String[] args)
{
int arr1[] = { 100, 3, 7, 50 };
int arr2[] = { 5, 1, 10, 4 };
int n = arr1.length;
int m = arr2.length;
System.out.println(totalPairs(arr1, arr2, n, m));
}
}
// This code is contributed by Rajput-Ji
Python 3
# Python3 program to count total number of
# pairs having elements with same sum of digits
# Function for returning
# sum of digits of a number
def digitSum(n):
Sum = 0
while n > 0:
Sum += n % 10
n = n // 10
return Sum
# Function to return the total pairs
# of elements with equal sum of digits
def totalPairs(arr1, arr2, n, m):
# set is used to avoid duplicate pairs
s = set()
for i in range(0, n):
for j in range(0, m):
# check sum of digits
# of both the elements
if digitSum(arr1[i]) == digitSum(arr2[j]):
if arr1[i] < arr2[j]:
s.add((arr1[i], arr2[j]))
else:
s.add((arr2[j], arr1[i]))
# return size of the set
return len(s)
# Driver code
if __name__ == "__main__":
arr1 = [100, 3, 7, 50]
arr2 = [5, 1, 10, 4]
n = len(arr1)
m = len(arr2)
print(totalPairs(arr1, arr2, n, m))
# This code is contributed by Rituraj Jain
C
// C# program to count total number of
// pairs having elements with same
// sum of digits
using System;
using System.Collections.Generic;
class GFG
{
public class pair
{
public int first, second;
public pair(int first, int second)
{
this.first = first;
this.second = second;
}
}
// Function for returning
// sum of digits of a number
static int digitSum(int n)
{
int sum = 0;
while (n > 0)
{
sum += n % 10;
n = n / 10;
}
return sum;
}
// Function to return the total pairs
// of elements with equal sum of digits
static int totalPairs(int []arr1, int []arr2,
int n, int m)
{
// set is used to avoid duplicate pairs
HashSet<pair> s = new HashSet<pair>();
for (int i = 0; i < n; i++)
{
for (int j = 0; j < m; j++)
{
// check sum of digits
// of both the elements
if (digitSum(arr1[i]) == digitSum(arr2[j]))
{
if (arr1[i] < arr2[j])
s.Add(new pair(arr1[i], arr2[j]));
else
s.Add(new pair(arr2[j], arr1[i]));
}
}
}
// return size of the set
return s.Count;
}
// Driver code
public static void Main(String[] args)
{
int []arr1 = { 100, 3, 7, 50 };
int []arr2 = { 5, 1, 10, 4 };
int n = arr1.Length;
int m = arr2.Length;
Console.WriteLine(totalPairs(arr1, arr2, n, m));
}
}
// This code is contributed by Princi Singh
java 描述语言
<script>
// Javascript program to count total number of
// pairs having elements with same
// sum of digits
// Function for returning
// sum of digits of a number
function digitSum(n)
{
var sum = 0;
while (n > 0) {
sum += n % 10;
n = parseInt(n / 10);
}
return sum;
}
// Function to return the total pairs
// of elements with equal sum of digits
function totalPairs(arr1, arr2, n, m)
{
// set is used to avoid duplicate pairs
var s = new Set();
for (var i = 0; i < n; i++) {
for (var j = 0; j < m; j++) {
// check sum of digits
// of both the elements
if (digitSum(arr1[i]) == digitSum(arr2[j])) {
if (arr1[i] < arr2[j])
s.add([arr1[i], arr2[j]]);
else
s.add([arr2[j], arr1[i]]);
}
}
}
// return size of the set
return s.size;
}
// Driver code
var arr1 = [100, 3, 7, 50 ];
var arr2 = [5, 1, 10, 4 ];
var n = arr1.length;
var m = arr2.length;
document.write( totalPairs(arr1, arr2, n, m));
</script>
Output:
3
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