diff --git a/algorithms/C/arrays/left_rotationOfarray.C b/algorithms/C/arrays/left_rotationOfarray.C
new file mode 100644
index 00000000..872b3a99
--- /dev/null
+++ b/algorithms/C/arrays/left_rotationOfarray.C
@@ -0,0 +1,61 @@
+/* PROBLEM: Given an array arr[] of integers, now make left rotation of the array according to the number given by user then print the updated array  */
+
+#include <stdio.h>
+#include<stdlib.h>
+void reverse(int*arr,int start,int end)
+{
+    while(start<end)
+    {
+        int temp;
+        temp=arr[start];
+        arr[start]=arr[end];
+        arr[end]=temp;
+        start++;
+        end--;
+    }
+}
+int main()
+{
+    int size;
+    printf("Enter size : "); // size is number of elements in the array
+    scanf("%d", &size);
+    int *arr;
+    arr=(int*)malloc(sizeof(int)*size);
+    for (int i = 0; i < size; i++)
+    {
+        scanf("%d", &arr[i]);
+    }
+    int no_of_rotation;
+    printf("Enter No Of left rotation:");
+    scanf("%d",&no_of_rotation);
+    if(size==1){
+       printf("%d",arr[0]);
+    }
+    else{
+        no_of_rotation=no_of_rotation%size; //fix the no of rotation ,if rotation number is gretter than size , then so that the no of rotation dose not excede the size .
+        reverse(arr,0,no_of_rotation-1);   //reverse the array 1st time toll the index of no_of_rotation-1
+        reverse(arr,no_of_rotation,size-1); //reverse the array elements left after the 1st time reverse.
+        reverse(arr,0,size-1);  //reverse the whole array once .
+    }
+    for(int i=0;i<size;i++)
+    {
+        printf("%d ",arr[i]);
+    }
+    
+   return 0;
+}
+
+/*
+Case 1:
+Input: Enter size: 5
+       1  2   3   4   5
+       Enter no_of_rotation:4
+Output:5 1 2 3 4
+Case 2:
+Input: Enter size: 6
+        7  9  5  6  3  2
+        Enter no_of_rotation:10
+Output: 3 2 7 9 5 6
+Time complexity: O(N) where N is the size of array
+*/
+