Divide array into two parts with equal sum according to the given constraints
Given an array arr[] of N integers, the task is to select an integer x (which may or may not be present in the array) and remove all of its occurrences from the array and divide the remaining array into two non-empty sub-sets such that:
- The elements of the first set are strictly smaller than x.
- The elements of the second set are strictly greater than x.
- The sum of the elements of both the sets is equal.
- If such an integer exists then print Yes otherwise print No.
Examples:
Input: arr[] = {1, 2, 2, 5}
Output: Yes
Choose x = 3, after removing all of its occurrences the array becomes arr[] = {1, 2, 2, 5}
{1, 2, 2} and {5} are the required sub-sets.Input: arr[] = {2, 1}
Output: No
Approach: The idea is to first sort the array and for all the numbers lying between 1 to maximum number present in the array, apply binary search and check if on removing all its occurrences from the array, sum of elements present on its left side (which are smaller than it) and sum of elements present on the right side (which are greater than it) is equal.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Function that checks if the given // conditions are satisfied void IfExists( int arr[], int n) { // To store the prefix sum // of the array elements int sum[n]; // Sort the array sort(arr, arr + n); sum[0] = arr[0]; // Compute the prefix sum array for ( int i = 1; i < n; i++) sum[i] = sum[i - 1] + arr[i]; // Maximum element in the array int max = arr[n - 1]; // Variable to check if there exists any number bool flag = false ; for ( int i = 1; i <= max; i++) { // Stores the index of the largest // number present in the array // smaller than i int findex = 0; // Stores the index of the smallest // number present in the array // greater than i int lindex = 0; int l = 0; int r = n - 1; // Find index of smallest number // greater than i while (l <= r) { int m = (l + r) / 2; if (arr[m] < i) { findex = m; l = m + 1; } else r = m - 1; } l = 1; r = n; flag = false ; // Find index of smallest number // greater than i while (l <= r) { int m = (r + l) / 2; if (arr[m] > i) { lindex = m; r = m - 1; } else l = m + 1; } // If there exists a number if (sum[findex] == sum[n - 1] - sum[lindex - 1]) { flag = true ; break ; } } // If no such number exists // print no if (flag) cout << "Yes" ; else cout << "No" ; } // Driver code int main() { int arr[] = { 1, 2, 2, 5 }; int n = sizeof (arr) / sizeof ( int ); IfExists(arr, n); return 0; } |
Java
// Java implementation of the approach import java.util.*; class GFG { // Function that checks if the given // conditions are satisfied static void IfExists( int arr[], int n) { // To store the prefix sum // of the array elements int sum[] = new int [n]; // Sort the array Arrays.sort(arr); sum[ 0 ] = arr[ 0 ]; // Compute the prefix sum array for ( int i = 1 ; i < n; i++) sum[i] = sum[i - 1 ] + arr[i]; // Maximum element in the array int max = arr[n - 1 ]; // Variable to check if there exists any number boolean flag = false ; for ( int i = 1 ; i <= max; i++) { // Stores the index of the largest // number present in the array // smaller than i int findex = 0 ; // Stores the index of the smallest // number present in the array // greater than i int lindex = 0 ; int l = 0 ; int r = n - 1 ; // Find index of smallest number // greater than i while (l <= r) { int m = (l + r) / 2 ; if (arr[m] < i) { findex = m; l = m + 1 ; } else r = m - 1 ; } l = 1 ; r = n; flag = false ; // Find index of smallest number // greater than i while (l <= r) { int m = (r + l) / 2 ; if (arr[m] > i) { lindex = m; r = m - 1 ; } else l = m + 1 ; } // If there exists a number if (sum[findex] == sum[n - 1 ] - sum[lindex - 1 ]) { flag = true ; break ; } } // If no such number exists // print no if (flag) System.out.println( "Yes" ); else System.out.println( "No" ); } // Driver code public static void main(String args[]) { int arr[] = { 1 , 2 , 2 , 5 }; int n = arr.length; IfExists(arr, n); } } // This code is contributed by Arnab Kundu |
Python3
# Python3 implementation of the approach # Function that checks if the given # conditions are satisfied def IfExists(arr, n) : # To store the prefix sum # of the array elements sum = [ 0 ] * n; # Sort the array arr.sort(); sum [ 0 ] = arr[ 0 ]; # Compute the prefix sum array for i in range ( 1 , n) : sum [i] = sum [i - 1 ] + arr[i]; # Maximum element in the array max = arr[n - 1 ]; # Variable to check if there # exists any number flag = False ; for i in range ( 1 , max + 1 ) : # Stores the index of the largest # number present in the array # smaller than i findex = 0 ; # Stores the index of the smallest # number present in the array # greater than i lindex = 0 ; l = 0 ; r = n - 1 ; # Find index of smallest number # greater than i while (l < = r) : m = (l + r) / / 2 ; if (arr[m] < i) : findex = m; l = m + 1 ; else : r = m - 1 ; l = 1 ; r = n; flag = False ; # Find index of smallest number # greater than i while (l < = r) : m = (r + l) / / 2 ; if (arr[m] > i) : lindex = m; r = m - 1 ; else : l = m + 1 ; # If there exists a number if ( sum [findex] = = sum [n - 1 ] - sum [lindex - 1 ]) : flag = True ; break ; # If no such number exists # print no if (flag) : print ( "Yes" ); else : print ( "No" ); # Driver code if __name__ = = "__main__" : arr = [ 1 , 2 , 2 , 5 ]; n = len (arr) ; IfExists(arr, n); # This code is contributed by Ryuga |
C#
// C# implementation of the approach using System; class GFG { // Function that checks if the given // conditions are satisfied static void IfExists( int [] arr, int n) { // To store the prefix sum // of the array elements int [] sum = new int [n]; // Sort the array Array.Sort(arr); sum[0] = arr[0]; // Compute the prefix sum array for ( int i = 1; i < n; i++) sum[i] = sum[i - 1] + arr[i]; // Maximum element in the array int max = arr[n - 1]; // Variable to check if there exists any number bool flag = false ; for ( int i = 1; i <= max; i++) { // Stores the index of the largest // number present in the array // smaller than i int findex = 0; // Stores the index of the smallest // number present in the array // greater than i int lindex = 0; int l = 0; int r = n - 1; // Find index of smallest number // greater than i while (l <= r) { int m = (l + r) / 2; if (arr[m] < i) { findex = m; l = m + 1; } else r = m - 1; } l = 1; r = n; flag = false ; // Find index of smallest number // greater than i while (l <= r) { int m = (r + l) / 2; if (arr[m] > i) { lindex = m; r = m - 1; } else l = m + 1; } // If there exists a number if (sum[findex] == sum[n - 1] - sum[lindex - 1]) { flag = true ; break ; } } // If no such number exists // print no if (flag) Console.WriteLine( "Yes" ); else Console.WriteLine( "No" ); } // Driver code public static void Main() { int [] arr = { 1, 2, 2, 5 }; int n = arr.Length; IfExists(arr, n); } } // This code is contributed by Code_Mech. |
PHP
<?php // PHP implementation of the approach // Function that checks if the given // conditions are satisfied function IfExists( $arr , $n ) { // To store the prefix $sum // of the array elements $sum = array_fill (0, $n , 0); // Sort the array sort( $arr ); $sum [0] = $arr [0]; // Compute the prefix sum array for ( $i = 1; $i < $n ; $i ++) $sum [ $i ] = $sum [ $i - 1] + $arr [ $i ]; // Maximum element in the array $max = $arr [ $n - 1]; // Variable to check if there exists any number $flag = false; for ( $i = 1; $i <= $max ; $i ++) { // Stores the index of the largest // number present in the array // smaller than i $findex = 0; // Stores the index of the smallest // number present in the array // greater than i $lindex = 0; $l = 0; $r = $n - 1; // Find index of smallest number // greater than i while ( $l <= $r ) { $m = ( $l + $r ) / 2; if ( $arr [ $m ] < $i ) { $findex = $m ; $l = $m + 1; } else $r = $m - 1; } $l = 1; $r = $n ; $flag = false; // Find index of smallest number // greater than i while ( $l <= $r ) { $m = ( $r + $l ) / 2; if ( $arr [ $m ] > $i ) { $lindex = $m ; $r = $m - 1; } else $l = $m + 1; } // If there exists a number if ( $sum [ $findex ] == $sum [ $n - 1] - $sum [ $lindex - 1]) { $flag = true; break ; } } // If no such number exists // print no if ( $flag == true) echo "Yes" ; else echo "No" ; } // Driver code $arr = array (1, 2, 2, 5 ); $n = sizeof( $arr ); IfExists( $arr , $n ); // This code is contributed by ihritik ?> |
Yes
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