Minimum count of increment of K size subarrays required to form a given Array | Function to find the minimum number of operations required to change all the array of zeros such that every element is greater than the given array ; Declaring the difference array of size N ; Number of operations ; First update the D [ i ] value with the previous value ; The index i has to be incremented ; We have to perform ( b [ i ] - d [ i ] ) operations more ; Increment the range i to i + k by need ; Check if i + k is valid index ; Driver code ; Function Call
function find_minimum_operations ( n , b , k ) { let d = new Array ( n + 1 ) ; d . fill ( 0 ) ; let i , operations = 0 , need ; for ( i = 0 ; i < n ; i ++ ) { if ( i > 0 ) { d [ i ] += d [ i - 1 ] ; } if ( b [ i ] > d [ i ] ) { operations += b [ i ] - d [ i ] ; need = b [ i ] - d [ i ] ; d [ i ] += need ; if ( i + k <= n ) { d [ i + k ] -= need ; } } } document . write ( operations ) ; } let n = 5 ; let b = [ 1 , 2 , 3 , 4 , 5 ] ; let k = 2 ; find_minimum_operations ( n , b , k ) ;