344. Reverse String and Reverse an Array in groups of given size

 

Write a function that reverses a string. The input string is given as an array of characters s.

You must do this by modifying the input array in-place with O(1) extra memory.

 

Example 1:

Input: s = ["h","e","l","l","o"]
Output: ["o","l","l","e","h"]

Example 2:

Input: s = ["H","a","n","n","a","h"]
Output: ["h","a","n","n","a","H"]

class Solution {

public void reverseString(char[] s) {

int left = 0;

int right = s.length - 1;

while(left < right){

char temp = s[left];

s[left] = s[right];

s[right] = temp;

left++;

right--;

}

}

}

 

 

Reverse an Array in groups of given size

 

Given an array arr[] and an integer k, find the array after reversing every subarray of consecutive k elements in place. If the last subarray has fewer than k elements, reverse it as it is. Modify the array in place, do not return anything.

Examples: 

Input: arr[] = [1, 2, 3, 4, 5, 6, 7, 8], k = 3
Output: [3, 2, 1, 6, 5, 4, 8, 7]
Explanation: Elements is reversed: [1, 2, 3] → [3, 2, 1], [4, 5, 6] → [6, 5, 4], and the last group [7, 8](size < 3) is reversed as [8, 7].

Input: arr[] = [1, 2, 3, 4, 5], k = 3
Output: [3, 2, 1, 5, 4]
Explanation: First group consists of elements 1, 2, 3. Second group consists of 4, 5.

Input: arr[] = [5, 6, 8, 9], k = 5
Output: [9, 8, 6, 5]
Explnation: Since k is greater than array size, the entire array is reversed.

 class gfg{

         public void revArrGroup(int[] nums,int k){

                int n = nums.length();

                for(int i = 0; i<n; i = i+k){

                    int left = i;

                    int right = Math.min(n-1,i+k-1)

                    while(left < right){

                            int temp = nums[left];

                            nums[left] = nums[right];

                           nums[right] = temp;

                           right--;left++;

                    }

                }

            }

    }