Left Rotate Array by K Positions

Solution with Approach by Abhilash

#include <stdio.h>

void rotate_left(int arr[], int n, int k) {
    // Your logic here
    if (n==0) return;
    k=k%n;
    if(k>n) return;
    int temp[k];
    for(int i=0; i<k; i++)
        temp[i] = arr[i];
    for(int i = 0; i<n-k; i++)
        arr[i] = arr[i+k];
    for(int i = n-k; i<n; i++)
        arr[i] = temp[i-n+k];
}

int main() {
    int n, k;
    scanf("%d %d", &n, &k);

    int arr[100];

    // Read array elements
    for (int i = 0; i < n; i++) {
        scanf("%d", &arr[i]);
    }

    // Rotate the array
    rotate_left(arr, n, k);

    // Print the rotated array
    for (int i = 0; i < n; i++) {
        printf("%d", arr[i]);
        if(i < n-1){
        	printf(" ");
        }
    }

    return 0;
}

Solving Approach

Edge cases

• if (n == 0) return;     // empty array, nothing to do
  k = k % n;              // normalize k (rotation beyond n is redundant)
  if (k > n) return;      // safety check (redundant after modulo)

Save the first k elements

• for (int i = 0; i < n - k; i++)
     arr[i] = arr[i + k];

Shift remaining elements left

• for (int i = 0; i < n - k; i++)
     arr[i] = arr[i + k];

• Example: [1,2,3,4,5] with k=2 becomes [3,4,5,4,5].

Restore saved elements at the end

• for (int i = n - k; i < n; i++)
     arr[i] = temp[i - n + k];

  Result: [3,4,5,1,2].