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Left Rotate an Array by K Places - Optimal Solution

Left Rotate an Array by K Places - Optimal Solution

Visualization

Algorithm Steps

  1. Given an array arr of size n and a value k.
  2. Normalize k using k = k % n to handle overflow.
  3. Reverse the first k elements.
  4. Reverse the remaining n - k elements.
  5. Reverse the entire array to get the final rotated form.

Left Rotate Array by K Places - Optimal Approach (Reversal Algorithm) Code

Python
JavaScript
Java
C++
C
def reverse(arr, start, end):
    while start < end:
        arr[start], arr[end] = arr[end], arr[start]
        start += 1
        end -= 1

def left_rotate(arr, k):
    n = len(arr)
    k = k % n
    reverse(arr, 0, k - 1)
    reverse(arr, k, n - 1)
    reverse(arr, 0, n - 1)
    return arr

# Sample Input
arr = [1, 2, 3, 4, 5, 6, 7]
k = 2
print("Rotated Array:", left_rotate(arr, k))

Detailed Step by Step Example

We will left rotate the array by 3 places using the reversal algorithm.

{ "array": [10,20,30,40,50,60,70], "showIndices": true }

Step 1: Reverse the first 3 elements.

{ "array": [10,20,30,40,50,60,70], "showIndices": true, "labels": { "0": "start", "2": "end" }, "highlightIndices": [0,1,2] }
{ "array": [30,20,10,40,50,60,70], "showIndices": true, "labels": { "0": "start", "2": "end" } }

Step 2: Reverse the remaining 4 elements.

{ "array": [30,20,10,40,50,60,70], "showIndices": true, "labels": { "3": "start", "6": "end" }, "highlightIndices": [3,4,5,6] }
{ "array": [30,20,10,70,60,50,40], "showIndices": true, "labels": { "3": "start", "6": "end" } }

Step 3: Reverse the entire array.

{ "array": [30,20,10,70,60,50,40], "showIndices": true, "labels": { "0": "start", "6": "end" }, "highlightIndices": [0,1,2,3,4,5,6] }
{ "array": [40,50,60,70,10,20,30], "showIndices": true, "labels": { "0": "start", "6": "end" } }

Final Rotated Array:

{ "array": [40,50,60,70,10,20,30], "showIndices": true, "highlightIndicesGreen": [4,5,6] }