In the world of algorithm design and problem-solving, the two-pointer technique stands out as a powerful and widely utilized strategy. With its application extending to arrays, linked lists, and strings, this approach has proven its efficacy in reducing time complexity and, in certain cases...
In this approach, we traverse the linked list using two pointers. One pointer is incremented by one while the other is incremented by two. When the fast pointer reaches the end, the slow pointer will be at the middle of the linked list. The time complexity of this solution is O(n...
How sorted array work? In a sorted array, if you want to find a pair satisfying some condition you can use this approach as at each step we know how to update the pointers. For better understanding, let's take an example. Suppose we have an array 'arr' and we want to find a pair...
Take a look at the implementation of the two-pointers approach: algorithm twoPointersMerge(A, B): // INPUT // A = The first array // B = The second array // n = the size of A // m = the size of B // OUTPUT // Returns the merged array of A and B i <- 0 j <- 0 R...
For every element you keep left and right positions for the element which is less than this element. (initially set to itselft). Iterate over descendant sorted array and update your info about pointers.
two pointers Approach : This implies that if there was a better solution possible, it will definitely have the Height greater than min(a1, aN). We know that, Base min(a1, aN) This means that we can discard min(a1, aN) from our set and look to solve this problem again from the ...
Using sorting/two pointers technique gives us a O(n*logn) time, and O(1) space(if quick sort is used) algorithm. This problem is different with the Two Sum problem in the following two aspects. 1. We need to design a data structure that supports adding new element to the internal ...
This is an important invariant that we must maintain for the correctness of this algorithm. Summarizing the above, Maintaining the invariant i + j = k - 1, If Bj-1 < Ai < Bj, then Ai must be the k-th smallest, or else if Ai-1 < Bj < Ai, then Bj must be the k-th ...
So our algorithm will guide us to move from any position after Y to Y finally. ——— b. If A[X] + A[Y] < 0: let a = |A[X] + A[Y]|, A[X] + A[Y+1] >= A[X] + A[Y] = -a with sorted array: If A[X] + A[Y+1] == -a ...
ListNode newNode = null; // carry to hold value if sum > 10 int carry = 0; // 2 pointers as we mentioned in the algorithm while (currentNodeList1 != null || currentNodeList2 != null) { // next iteration might bring a carry with it, which we need to be added into the sum. ...