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Count of suffix increment/decrement operations to construct a given array in C++
We are given a target array arr[] containing positive integers. The goal is to construct the target array arr[] using an initial array with all 0s. The operations that can be applied on a given empty array with all 0s will suffix increment/decrement operations.
If we choose any index say i, then in case of suffix increment operation we will add 1 to all elements from index i till last index.
In case of suffix decrement operation we will subtract 1 from all elements from index i till last index.
Let us understand with examples
Input − arr[]= { 1,2,3 }
Output − Count of suffix increment/decrement operations to construct a given array are − 3
Explanation
Starting from { 0, 0, 0 }
Choose index 0, applying suffix increment { 1, 1, 1 }
Choose index 1, applying suffix increment { 1, 2, 2 }
Choose index 2, applying suffix increment { 1, 2, 3 }
Total operations =3Input − arr[]= { 1, 4, 5, 3 }
Output − Count of suffix increment/decrement operations to construct a given array are − 7
Explanation
Starting from { 0, 0, 0, 0 }
Choose index 0, applying suffix increment { 1, 1, 1, 1 }
Choose index 1, applying suffix increment { 1, 2, 2, 2 }
Choose index 1, applying suffix increment { 1, 3, 3, 3 }
Choose index 1, applying suffix increment { 1, 4, 4, 4 }
Choose index 2, applying suffix increment { 1, 4, 5, 5 }
Choose index 3, applying suffix decrement { 1, 4, 5, 4 }
Choose index 3, applying suffix decrement { 1, 4, 5, 3 }
Total operations = 7Approach used in the below program is as follows
If we take the initial array as B[ ]. To make first element B[0] equal to arr[0]. We will need arr[0] number of suffix increment operations. After this all B[0]=B[1]....=B[n-1]=arr[0] will be equal.
To make second element B[1] equal to arr[1], we will need | arr[1]-arr[0] | number of operations. (either increment or decrement).
So to make B[i] equal to arr[i] we will need | arr[i]- arr[i-1] | number of operations.
Total count of operations will be | arr[0] | + | arr[1]-arr[0] | + …. + | arr[n-1]-arr[n-2] |.
Take the target array as arr[].
Function incr_decr_op(int arr[], int size) takes array and its length and returns the count of suffix increment/decrement operations to construct a given array
Take the initial count as 0.
Traverse array arr[] using for loop
For index 0 add arr[i] to count.
For other indexes add abs( arr[i]-arr[i-1] ) to count.
At the end of the for loop return count as result.
Example
#include <bits/stdc++.h>
using namespace std;
int incr_decr_op(int arr[], int size){
int count = 0;
for (int i = 0; i < size; i++){
if (i > 0){
count += abs(arr[i] - arr[i - 1]);
}
else{
count = count + abs(arr[i]);
}
}
return count;
}
int main(){
int arr[] = { 3, 3, 1, 2, 2 };
int size = sizeof(arr) / sizeof(arr[0]);
cout<<"Count of suffix increment/decrement operations to construct a given array are: "<<incr_decr_op(arr, size) << endl;
}Output
If we run the above code it will generate the following output −
Count of suffix increment/decrement operations to construct a given array are: 6
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