Minimize prize count required such that smaller value gets less prize in an adjacent pair | C ++ implementation to find the minimum prizes required such that adjacent smaller elements gets less number of prizes ; Function to find the minimum number of required such that adjacent smaller elements gets less number of prizes ; Loop to compute the smaller elements at the left ; Loop to find the smaller elements at the right ; Loop to find the minimum prizes required ; Driver Code
#include <bits/stdc++.h> NEW_LINE using namespace std ; int minPrizes ( int arr [ ] , int n ) { int dpLeft [ n ] ; dpLeft [ 0 ] = 1 ; for ( int i = 1 ; i < n ; i ++ ) { if ( arr [ i ] > arr [ i - 1 ] ) { dpLeft [ i ] = dpLeft [ i - 1 ] + 1 ; } else { dpLeft [ i ] = 1 ; } } int dpRight [ n ] ; dpRight [ n - 1 ] = 1 ; for ( int i = n - 2 ; i >= 0 ; i -- ) { if ( arr [ i ] > arr [ i + 1 ] ) { dpRight [ i ] = dpRight [ i + 1 ] + 1 ; } else { dpRight [ i ] = 1 ; } } int totalPrizes = 0 ; for ( int i = 0 ; i < n ; i ++ ) { totalPrizes += max ( dpLeft [ i ] , dpRight [ i ] ) ; } cout << totalPrizes << endl ; return 0 ; } int main ( ) { int arr [ ] = { 1 , 2 , 2 , 3 } ; int n = sizeof ( arr ) / sizeof ( arr [ 0 ] ) ; minPrizes ( arr , n ) ; return 0 ; }