Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
Count valid pairs in the array satisfying given conditions in C++
We are given with an array arr[] of N elements. The goal is to find the count of all valid pairs (Arr[i],Arr[j]) that follow certain conditions. Pairs Arr[i],Arr[j] invalid if −
- Arr[i]==Arr[j]
- Arr[i]+Arr[j] is even
- i+j<120
Note − Arr[i],Arr[j] and Arr[j],Arr[i] will be counted as one pair. Valid pair has i!=j Let’s understand with examples.
Input
Arr[]= { 3,2,1,2,4,3 } N=4Output
Count of valid pairs: 2
Explanation
Valid pairs are −
Arr[0] & Arr[4] → (3,3) here Arr[i]==Arr[j] & 3+3 is even also i!=j and i+j<120 Arr[1] & Arr[3] → (2,2) here Arr[i]==Arr[j] & 2+2 is even also i!=j and i+j<120
Input
Arr[]= { 1,2,3,4,5,6 } N=6Output
Count of valid pairs: 0
Explanation
No repetition of elements. No pair of type ( a,a ) possible where i!=j.
Approach used in the below program is as follows
- We take an integer array Arr[] initialized with random numbers.
- Take a variable n which stores the length of Arr[].
- Function countPairs(int arr[], int n) takes an array, its length as input and returns the pairs which are valid and meet desired conditions.
- Traverse array using two for loops for each element of the pair.
- Outer Loop from 0<=i<n-1, inner loop i<j<n
- Calculate sum of arr[i], arr[j] assum=(arr[i]+aar[j]).
- To check if a pair is valid. Compare if i!=j and sum%2==0 also i+j<120.
- Now check if arr[i]==arr[j]. Increment count.
- At the end of all loops count will have a total number of pairs that are valid
- Return the count as result.
Example
#include <bits/stdc++.h>
using namespace std;
int countPairs(int arr[], int n){
int count=0;
for(int i=0;i<n;i++){
for(int j=i+1;j<n;j++) //j=i+1 we don't have to check i!=j{
int sum=arr[i]+arr[j];
//valid pairs have i!=j
if( sum%2==0 && i+j<120){
if( arr[i]==arr[j] ) //valid pair{
count++;
cout<<endl<<" a:"<<arr[i]<<"b: "<<arr[j];
}
}
}
}
return count;
}
int main(){
int arr[] = {1,2,3,2,4,1,4 };
int n = sizeof(arr) / sizeof(arr[0]);
cout <<endl<<"Valid pairs in array:"<<countPairs(arr, n);
return 0;
}Output
Valid pairs in array: a:1b: 1 a:2b: 2 a:4b: 43
Updated on: 16-Sep-2020
370 Views
Advertisements