#include <stdio.h>
void findPair(int nums[], int n, int target)
{
// consider each element except the last
for (int i = 0; i < n - 1; i++)
{
// start from the i'th element until the last element
for (int j = i + 1; j < n; j++)
{
// if the desired sum is found, print it
if (nums[i] + nums[j] == target)
{
printf("Pair found (%d, %d)
", nums[i], nums[j]);
return;
}
}
}
// we reach here if the pair is not found
printf("Pair not found");
}
int main(void)
{
int nums[] = { 8, 7, 2, 5, 3, 1 };
int target = 10;
int n = sizeof(nums)/sizeof(nums[0]);
findPair(nums, n, target);
return 0;
}
#include <iostream>
#include <unordered_map>
using namespace std;
// Function to find a pair in an array with a given sum using hashing
void findPair(int nums[], int n, int target)
{
// create an empty map
unordered_map<int, int> map;
// do for each element
for (int i = 0; i < n; i++)
{
// check if pair (nums[i], target - nums[i]) exists
// if the difference is seen before, print the pair
if (map.find(target - nums[i]) != map.end())
{
cout << "Pair found (" << nums[map[target - nums[i]]] << ", "
<< nums[i] << ")
";
return;
}
// store index of the current element in the map
map[nums[i]] = i;
}
// we reach here if the pair is not found
cout << "Pair not found";
}
int main()
{
int nums[] = { 8, 7, 2, 5, 3, 1 };
int target = 10;
int n = sizeof(nums)/sizeof(nums[0]);
findPair(nums, n, target);
return 0;
}
#include <iostream>
#include <algorithm>
using namespace std;
// Function to find a pair in an array with a given sum using sorting
void findPair(int nums[], int n, int target)
{
// sort the array in ascending order
sort(nums, nums + n);
// maintain two indices pointing to endpoints of the array
int low = 0;
int high = n - 1;
// reduce the search space `nums[low…high]` at each iteration of the loop
// loop till the search space is exhausted
while (low < high)
{
// sum found
if (nums[low] + nums[high] == target)
{
cout << "Pair found (" << nums[low] << ", " << nums[high] << ")
";
return;
}
// increment `low` index if the total is less than the desired sum;
// decrement `high` index if the total is more than the desired sum
(nums[low] + nums[high] < target)? low++: high--;
}
// we reach here if the pair is not found
cout << "Pair not found";
}
int main()
{
int nums[] = { 8, 7, 2, 5, 3, 1 };
int target = 10;
int n = sizeof(nums)/sizeof(nums[0]);
findPair(nums, n, target);
return 0;
}
class Main
{
// Naive method to find a pair in an array with a given sum
public static void findPair(int[] nums, int target)
{
// consider each element except the last
for (int i = 0; i < nums.length - 1; i++)
{
// start from the i'th element until the last element
for (int j = i + 1; j < nums.length; j++)
{
// if the desired sum is found, print it
if (nums[i] + nums[j] == target)
{
System.out.println("Pair found (" + nums[i] + "," + nums[j] + ")");
return;
}
}
}
// we reach here if the pair is not found
System.out.println("Pair not found");
}
public static void main (String[] args)
{
int[] nums = { 8, 7, 2, 5, 3, 1 };
int target = 10;
findPair(nums, target);
}
}
Counting Pairs in an Array, resulting in a given sum
<script>
/* javascript implementation of simple method to find count of
pairs with given sum*/
var arr = [ 1, 5, 7, -1, 5 ];
// Returns number of pairs in arr[0..n-1] with sum equal
// to 'sum'
function getPairsCount(n , sum) {
var hm = new Map();
// Store counts of all elements in map hm
for (var i = 0; i < n; i++) {
// initializing value to 0, if key not found
if (!hm.has(arr[i]))
hm.set(arr[i], 0);
hm.set(arr[i], hm.get(arr[i]) + 1);
}
var twice_count = 0;
// iterate through each element and increment the
// count (Notice that every pair is counted twice)
for (i = 0; i < n; i++) {
if (hm.get(sum - arr[i]) != null)
twice_count += hm.get(sum - arr[i]);
// if (arr[i], arr[i]) pair satisfies the
// condition, then we need to ensure that the
// count is decreased by one such that the
// (arr[i], arr[i]) pair is not considered
if (sum - arr[i] == arr[i])
twice_count--;
}
// return the half of twice_count
return twice_count / 2;
}
// Driver method to test the above function
var sum = 6;
document.write("Count of pairs is " + getPairsCount(arr.length, sum));
// This code is contributed by umadevi9616
</script>
Function to find a pair in an array with a given sum using hashing
def findPair(A, sum):
# create an empty dictionary
dict = {}
# do for each element
for i, e in enumerate(A):
# check if pair `(e, sum-e)` exists
# if the difference is seen before, print the pair
if sum - e in dict:
print("Pair found at index", dict.get(sum - e), "and", i)
return
# store index of the current element in the dictionary
dict[e] = i
# No pair with the given sum exists in the list
print("Pair not found")
Counting Pairs in an Array, resulting in a given sum
<script>
// javascript implementation of simple method to find count of
// pairs with given sum.
// Returns number of pairs in arr[0..n-1] with sum equal
// to 'sum'
function getPairsCount(arr , n , k) {
var m = new Map();
var count = 0;
for (var i = 0; i < n; i++) {
if (m.has(k - arr[i])) {
count += m.get(k - arr[i]);
}
if (m.has(arr[i])) {
m.set(arr[i], m.get(arr[i]) + 1);
} else {
m.set(arr[i], 1);
}
}
return count;
}
// Driver function to test the above function
var arr = [ 1, 5, 7, -1, 5 ];
var n = arr.length;
var sum = 6;
document.write("Count of pairs is " + getPairsCount(arr, n, sum));
// This code is contributed by umadevi9616
</script>
find a pair of elements from an array whose sum equals a given number python
#Given an array of integers and a number, write a function that finds two elements from the array whose sum is equal to the given number.
def find_pair(arr, num):
for i in range(len(arr)):
for j in range(i + 1, len(arr)):
if arr[i] + arr[j] == num:
return [arr[i], arr[j]]
print(find_pair([3, 6, 8, -8, 10, 8], 16))
[6, 10]