count = 0
given_sum = 7
x = 4
y = 3
pair = x + y
if pair == 7:
count += 1
#count goes up if the sum of x and y is the given number<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><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>