# finding three elements in an array whose sum is equal to a given number

Finding three elements in an array whose sum is equal to a given number(say trisum)

Here, for every element(arr[i]) that we pick, trisum-arr[i] gives us the bisum (algorithm to find a pair of elements that matches to sum k), hence we make use of our bisum algorithm to find a pair that matches to sum k

Assumption: The input array is sorted.

Example
arr: {1,2,3,4,5} and trisum is 8 then tri pairs are 1,3,4 and 1,2,5
Here lets say arr[i] is 1 then bisum=trisum-arr[i] i.e, bisum=8-1=7 therefore we should run bisum algorithm for sum 7

Time complexity for this algo: O(n^2) because for every element we need to find set of pairs

Note: If the array is not sorted the the complexity would be nlogn + O(n^2)

```/**
* find 3 elements in an array, that sum up to particular number with best
* complexity
*
* @author ntallapa
*
*/
public class TriPairSumCloseToK {

public void getTriPair(int[] arr, int expectedTriSum) {
for (int i = 0; i < arr.length; i++) {
int expectedBiSum = expectedTriSum - arr[i];
String output = getPair2(arr, expectedBiSum, i);
if(!output.equalsIgnoreCase("")) {
System.out.println(arr[i]+","+output.toString());
}
}
}

/**
* Find the pair in an array whose sum with complexity O(n). This assumes
* the array passed is sorted array and there are no duplicates in the
* array
* @param arr
*            input array of elements
* @param expectedBiSum
*            sum for which pair of array elements needs to be searched
*/
public String getPair2(int[] arr, int expectedBiSum, int ignoreIndex) {
int start = 0;
int end = arr.length - 1;
int sum = 0;

// output array to record matching pairs
StringBuilder arrs = new StringBuilder();

while (start < end) {
//
if(start == ignoreIndex)
start++;
//
if(end == ignoreIndex)
end--;

sum = arr[start] + arr[end];
if (sum == expectedBiSum) {
// we have found one pair, add it to our output array
arrs.append(arr[start] + "," + arr[end] + ";");
start++;
end--;
} else if (sum < expectedBiSum) {
// if the sum of the pair is less than the sum we are searching
// then increment the start pointer which would give us the sum
// more than our current sum as the array is sorted
start++;
} else {
// if the sum of the pair is greater than the sum we are
// searching then decrement the end pointer which would give us
// the sum less than our current sum as the array is sorted
end--;
}
}
return arrs.toString();
}

/**
* Client to test
*
* @param args
*/
public static void main(String[] args) {
TriPairSumCloseToK p = new TriPairSumCloseToK();

// sorted array algo
int[] arr1 = { 1, 2, 3, 4, 5 };
p.getTriPair(arr1, 1);

}
}
```

### 3 Responses to finding three elements in an array whose sum is equal to a given number

1. hilton6l1t says:

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• Niranjan says:

thank you hilton6l1t

2. Satyajit says:

public static void findPairs(int arr[] , int n){
int r[] = new int;
for(int i=0; i<arr.length ;i++){
if(arr[i]< n){
r=arr[i];
for(int j=0; j<arr.length && (j != i) ; j++ ){
if(r+arr[j] < n){
r=arr[j];
for(int k=0; k<arr.length &&( k != i ) && ( k!= j);k++){
if(r+r+arr[k] == n){
r= arr[k];
System.out.println("pair :"+ r+" "+r+" "+r);
}
}
}
}
}
}
}

public static void main(String[] args) {
int arr[] = {2,4,8,1,6,9,3 ,1 };
findPairs(arr , 16 );
}

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