POTD_27_OCT_2024_TripletFamily #214
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Efficient approach: The idea is similar to Find a triplet that sum to a given value. Sort the given array first. Start fixing the greatest element of three from the back and traverse the array to find the other two numbers which sum up to the third element. Take two pointers j(from front) and k(initially i-1) to find the smallest of the two number and from i-1 to find the largest of the two remaining numbers If the addition of both the numbers is still less than A[i], then we need to increase the value of the summation of two numbers, thereby increasing the j pointer, so as to increase the value of A[j] + A[k]. If the addition of both the numbers is more than A[i], then we need to decrease the value of the summation of two numbers, thereby decrease the k pointer so as to decrease the overall value of A[j] + A[k]. Time Complexity: O(N^2) Auxiliary Space: O(1)
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Gyanthakur
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Oct 31, 2024
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Efficient approach: The idea is similar to Find a triplet that sum to a given value.
Sort the given array first.
Start fixing the greatest element of three from the back and traverse the array to find the other two numbers which sum up to the third element.
Take two pointers j(from front) and k(initially i-1) to find the smallest of the two number and from i-1 to find the largest of the two remaining numbers
If the addition of both the numbers is still less than A[i], then we need to increase the value of the summation of two numbers, thereby increasing the j pointer, so as to increase the value of A[j] + A[k].
If the addition of both the numbers is more than A[i], then we need to decrease the value of the summation of two numbers, thereby decrease the k pointer so as to decrease the overall value of A[j] + A[k]. Time Complexity: O(N^2)
Auxiliary Space: O(1)
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Fixes: #[issue_number] (replace with the issue number, if applicable)
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