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Maximize the number of sum pairs which are divisible by K in C++
Given the task is to calculate the maximum number of pairs arr[i] + arr[j] that are divisible by K where arr[] is an array containing N integers.
Given the condition that a particular index number cannot be used in more than one pair.
Input
arr[]={1, 2 ,5 ,8 ,3 }, K=2Output
2
Explanation − The desired pairs are: (0,2), (1,3) as 1+5=6 and 2+8=10 . Both 6 and 10 are divisible by 2.
Alternative answers could be the pairs: (0,4), (1,3) or (2,4), (1,3) but the answer remains the same, that is, 2.
Input
arr[]={1 ,3 ,5 ,2 ,3 ,4 }, K=3Output
3
Approach used in the below program as follows
In the variable n of type int store the size of the array
In the function MaxPairs() use unordered map and increase the value as um[arr[i]%K] by one for each element of the array.
Iterate and get every possible um value.
If the um value is zero then the number of pairs will become um[0]/2.
Then pairs can be formed from every um value ‘a’ by using the minimum of (um[a], um[Ka])
Subtract from the um value, the number of pairs used.
Example
#include <bits/stdc++.h>
using namespace std;
int MaxPairs(int arr[], int size, int K){
unordered_map<int, int> um;
for (int i = 0; i < size; i++){
um[arr[i] % K]++;
}
int count = 0;
/*Iteration for all the number that are less than the hash value*/
for (auto it : um){
// If the number is 0
if (it.first == 0){
//Taking half since same number
count += it.second / 2;
if (it.first % 2 == 0){
um[it.first] = 0;
}
else{
um[it.first] = 1;
}
}
else{
int first = it.first;
int second = K - it.first;
// Check for minimal occurrence
if (um[first] < um[second]){
//Taking the minimal
count += um[first];
//Subtracting the used pairs
um[second] -= um[first];
um[first] = 0;
}
else if (um[first] > um[second]){
// Taking the minimal
count += um[second];
//Subtracting the used pairs
um[first] -= um[second];
um[second] = 0;
}
else{
//Checking if the numbers are same
if (first == second){
//If same then number of pairs will be half
count += it.second / 2;
//Checking for remaining
if (it.first % 2 == 0)
um[it.first] = 0;
else
um[it.first] = 1;
}
else{
//Storing the number of pairs
count += um[first];
um[first] = 0;
um[second] = 0;
}
}
}
}
return count;
}
//Main function
int main(){
int arr[] = { 3, 6, 7, 9, 4, 4, 10 };
int size = sizeof(arr) / sizeof(arr[0]);
int K = 2;
cout<<"Maximize the number of sum pairs which are divisible by K is: "<<MaxPairs(arr, size, K);
return 0;
}Output
If we run the above code, we will get the following output −
Maximize the number of sum pairs which are divisible by K is: 3
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