Javascript Program to Find a triplet such that sum of two equals to third element
Last Updated :
20 Apr, 2023
Given an array of integers, you have to find three numbers such that the sum of two elements equals the third element.
Examples:
Input: {5, 32, 1, 7, 10, 50, 19, 21, 2}
Output: 21, 2, 19
Input: {5, 32, 1, 7, 10, 50, 19, 21, 0}
Output: no such triplet exist
Question source: Arcesium Interview Experience | Set 7 (On campus for Internship)
Simple approach: Run three loops and check if there exists a triplet such that sum of two elements equals the third element.
Time complexity: O(n^3)
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].
Below image is a dry run of the above approach:
Below is the implementation of the above approach:
Javascript
<script>
function findTriplet(arr, n)
{
arr.sort((a,b) => a-b);
for (let i = n - 1; i >= 0; i--)
{
let j = 0;
let k = i - 1;
while (j < k)
{
if (arr[i] == arr[j] + arr[k])
{
document.write( "numbers are " + arr[i] +
" " + arr[j] + " " +
arr[k] + "<br>" );
return ;
}
else if (arr[i] > arr[j] + arr[k])
j += 1;
else
k -= 1;
}
}
document.write( "No such triplet exists" );
}
let arr = [5, 32, 1, 7, 10, 50, 19, 21, 2];
let n = arr.length;
findTriplet(arr, n);
</script>
|
Output:
numbers are 21 2 19
Time complexity: O(N^2)
Space Complexity: O(1) as no extra space has been used.
Another Approach: The idea is similar to previous approach.
- Sort the given array.
- Start a nested loop, fixing the first element i(from 0 to n-1) and moving the other one j (from i+1 to n-1).
- Take the sum of both the elements and search it in the remaining array using Binary Search.
Javascript
<script>
bool search(sum, start, end, arr)
{
while (start <= end)
{
let mid = (start + end) / 2;
if (arr[mid] == sum)
{
return true ;
}
else if (arr[mid] > sum)
{
end = mid - 1;
}
else
{
start = mid + 1;
}
}
return false ;
}
function findTriplet(arr, n)
{
arr.sort((a,b) => a-b);
for (let i = 0; i < n; i++)
{
for (let j = i + 1; j < n; j++)
{
if (search((arr[i] + arr[j]),
j, n - 1, arr))
{
document.write( "numbers are " + arr[i] +
" " + arr[j] + " " +
( arr[i] + arr[j] ) + "<br>" );
}
}
}
document.write( "No such triplet exists" );
}
let arr = [5, 32, 1, 7, 10, 50, 19, 21, 2];
let n = arr.length;
findTriplet(arr, n);
</script>
|
Time Complexity: O(N^2*log N)
Space Complexity: O(1)
Please refer complete article on Find a triplet such that sum of two equals to third element for more details!
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