Problem Statement:
You are given an array A containing N elements. You have to reverse this array using constant space and an O(N) time complexity. This means you are not allowed to declare another array of the same size to store the reversed elements. You must reverse the elements in the same array, modifying it in place.
Approach:
The key to solving this problem efficiently is using two pointers to swap the elements in place.
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Two pointers: You initialize two pointers, l (left) starting at index 0 and r (right) starting at index N - 1.
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Swapping elements: You swap the values of A[l] and A[r]. This means the element at the left end of the array will swap with the element at the right end.
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Move inward: After each swap, increment l (move it towards the right) and decrement r (move it towards the left).
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Repeat: Continue swapping until l is no longer less than r. This ensures that all elements have been reversed in the same array without using extra space.
The key point is that since you’re only moving two pointers and swapping elements, the array is reversed in place, and you only traverse the array once, achieving O(N) time complexity.
Time Complexity:
- The time complexity is O(N), where N is the size of the array. This is because you need to iterate through half the array, but every swap involves two elements, ensuring linear time complexity.
Space Complexity:
- The space complexity is O(1), meaning the algorithm uses constant space, as you are not using any additional arrays or large data structures, just a few variables like l and r.