![]() If they do not, we will do one of two things: At every step, we will see if the numbers pointed by the two pointers add up to the target sum. Given that the input array is sorted, an efficient way would be to start with one pointer in the beginning and another pointer at the end. The time complexity of this algorithm will be O ( N 2 ) O(N^2) O ( N 2 ) where ‘N’ is the number of elements in the input array. To solve this problem, we can consider each element one by one (pointed out by the first pointer) and iterate through the remaining elements (pointed out by the second pointer) to find a pair with the given sum. Given an array of sorted numbers and a target sum, find a pair in the array whose sum is equal to the given target. For example, take a look at the following problem: The set of elements could be a pair, a triplet or even a subarray. In problems where we deal with sorted arrays (or LinkedLists) and need to find a set of elements that fulfill certain constraints, the Two Pointers approach becomes quite useful. ![]()
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