Question

Let's say someone gives you an array that is filled with P numbers. Please next see if there are two numbers whose sum equals a given number S and determine if there is two numbers that do this. Take for example, if I give you the input array to be 8, 2, 1, 7, and our variable S is 15, then the answer would be yes because 8 and 7 add up to S which in this case is 15. You are allowed to use a number twice. To solve this problem, please  Give me an O(NlogN) algorithm. (In language C++)

Answer 1

Code:

#include<iostream>
using namespace std;
bool check_pair(int *arr,int n,int sum){
         int first=0,last=n-1;
   while(first<last){
        if (arr[first]+arr[last]<sum)
                first++;
        else if (arr[first]+arr[last]>sum)
        last--;
        else return 1;
   }
}
void swapping(int &a, int &b) {     //swap the content of a and b
   int temp;
   temp = a;
   a = b;
   b = temp;
}
void display(int *array, int size) {
   for(int i = 0; i<size; i++)
      cout << array[i] << " ";
   cout << endl;
}
void merge(int *array, int l, int m, int r) {
   int i, j, k, nl, nr;
   //size of left and right sub-arrays
   nl = m-l+1; nr = r-m;
   int larr[nl], rarr[nr];
   //fill left and right sub-arrays
   for(i = 0; i<nl; i++)
      larr[i] = array[l+i];
   for(j = 0; j<nr; j++)
      rarr[j] = array[m+1+j];
   i = 0; j = 0; k = l;
   //marge temp arrays to real array
   while(i < nl && j<nr) {
      if(larr[i] <= rarr[j]) {
         array[k] = larr[i];
         i++;
      }else{
         array[k] = rarr[j];
         j++;
      }
      k++;
   }
   while(i<nl) {       //extra element in left array
      array[k] = larr[i];
      i++; k++;
   }
   while(j<nr) {     //extra element in right array
      array[k] = rarr[j];
      j++; k++;
   }
}
void mergeSort(int *array, int l, int r) {
   int m;
   if(l < r) {
      int m = l+(r-l)/2;
      // Sort first and second arrays
      mergeSort(array, l, m);
      mergeSort(array, m+1, r);
      merge(array, l, m, r);
   }
}
int main() {
   int n;
   cout << "Enter the number of elements: ";
   cin >> n;
   int arr[n];     //create an array with given number of elements
   cout << "Enter elements:" << endl;
   for(int i = 0; i<n; i++) {
      cin >> arr[i];
   }
   cout << "Array before Sorting: ";
   display(arr, n);
   mergeSort(arr, 0, n-1);     //(n-1) for last index
   cout << "Array after Sorting: ";
   display(arr, n);
   //solution for your question starts here.
   printf("enter the sum for which you want to find pair\n");
   int sum;
   scanf("%d",&sum);
   bool con=check_pair(arr,n,sum);
   if (con)
        cout<<"pair available\n";
   else
        cout<<"pair unavailable\n";
   return 0;

}

Explanation:

Algorithm:

1.sort array (merge sort o(n log n)

2. Take two pointers first and last.

3. while(first<last)

do{

if arr[first]+arr[last]<sum

first++

else if arr[first]+arr[last]>sum

last--

else

return 1

}--->o(n)

This whole program takes total complexity o(nlogn)

Hope it helps.

I hope you will accept my answer