701191. Reverse array in groups
Reverse array in groups0255
Summary
Reverse an array in groups, where each group contains a specified number of elements. This problem requires us to understand how arrays work and how we can manipulate them efficiently.
Detailed Explanation
Given an integer n, representing the size of the array, and another integer k, representing the size of each group, our task is to reverse every group in the array. For example, if we have an array [1, 2, 3, 4, 5, 6, 7] with n = 7 and k = 3, the output should be [3, 2, 1, 7, 6, 5, 4].
To solve this problem, we can iterate through the array in groups. For each group, we reverse the elements before moving on to the next group. This approach ensures that every group is reversed correctly.
Here's a step-by-step breakdown of the solution:
- Initialize two pointers,
startandend, to represent the current group. - Iterate through the array until
endexceeds the length of the array or the end of the current group is reached. - During each iteration, if the current element belongs to the current group (i.e., its index is a multiple of
k), reverse the elements betweenstartandend. - Update
startandendfor the next group. - Repeat steps 2-4 until the end of the array is reached.
The time complexity of this solution is O(n), where n is the size of the array, since we're iterating through the array once. The space complexity is O(1) because we're not using any additional space that scales with the input size.
Optimized Solutions
Java
public void reverseArrayInGroups(int[] arr, int k) {
for (int i = 0; i < arr.length; i += k) {
int start = Math.min(i, arr.length - 1);
int end = Math.min(start + k - 1, arr.length - 1);
int left = start;
int right = end;
while (left < right) {
int temp = arr[left];
arr[left] = arr[right];
arr[right] = temp;
left++;
right--;
}
}
}