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Count number of 1s in the array after N moves in C
We are given an array of size N. The array has all 0’s initially. The task is to count the no. of 1’s in the array after N moves. Each Nth move has a rule associated. The rules are −
1st Move − Change element at positions 1, 2, 3, 4…………..
2nd Move − Change element at positions 2, 4, 6, 8…………..
3rd Move − Change element at positions 3, 6, 9, 12…………..
Count the number of 1’s in the last array.
Let’s understand with examples.
Input
Arr[]={ 0,0,0,0 } N=4Output
Number of 1s in the array after N moves − 2
Explanation − Array after following moves −
Move 1: { 1,1,1,1 }
Move 2: { 1,0,1,0 }
Move 3: { 1,0,0,3 }
Move 4: { 1,0,0,1 }
Number of ones in the final array is 2.Input
Arr[]={ 0,0,0,0,0,0} N=6Output
Number of 1s in the array after N moves − 2
Explanation − Array after following moves −
Move 1: { 1,1,1,1,1,1,1 }
Move 2: { 1,0,1,0,1,0,1 }
Move 3: { 1,0,0,1,0,0,1 }
Move 4: { 1,0,0,0,1,0,0 }
Move 5: { 1,0,0,0,0,1,0 }
Move 4: { 1,0,0,0,0,0,1 }
Number of ones in the final array is 2.Approach used in the below program is as follows
We take an integer array Arr[] initialized with 0’s and integer N.
Function Onecount takes Arr[] and it’s size N as input and returns no. of ones in the final array after N moves.
The for loop starts from 1 till the end of the array.
Each i represents the ith move.
Nested for loop starts from 0th index till end of array.
For each ith move, if the index j is multiple of i (j%i==0), replace the 0 with 1 at that position.
This process continues for each i till the end of array.
Note − Index starts from i=1,j=1 but array indexes are from 0 to N-1. For this reason arr[j1] is converted each time.
Finally traverse the whole array again and count no. of 1’s in it and store in count.
- Return count as desired result.
Example
#include <stdio.h>
int Onecount(int arr[], int N){
for (int i = 1; i <= N; i++) {
for (int j = i; j <= N; j++) {
// If j is divisible by i
if (j % i == 0) {
if (arr[j - 1] == 0)
arr[j - 1] = 1; // Convert 0 to 1
else
arr[j - 1] = 0; // Convert 1 to 0
}
}
}
int count = 0;
for (int i = 0; i < N; i++)
if (arr[i] == 1)
count++; // count number of 1's
return count;
}
int main(){
int size = 6;
int Arr[6] = { 0 };
printf("Number of 1s in the array after N moves: %d", Onecount(Arr, size));
return 0;
}Output
If we run the above code it will generate the following output −
Number of 1s in the array after N moves: 2
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