How to find all ordered pairs of elements in array of integers whose sum lies in a given range of value

Question

Given an array of integers find the number of all ordered pairs of elements in the array whose sum lies in a given range [a,b]

Here is an O(n^2) solution for the same <

Answer 1

Sort the array first and count the pairs by two indexes. The two indexes approach is similar to the one in 2-sum problem, which avoids the binary-search for N times. The time consuming of the algorithm is Sort Complexity + O(N), typically, sort is O(NlnN), thus this approach is O(NlnN). The idea of the algorithm is, for an index i, find an lower bound and an upper bound such that a <= arr[i]+arr[low] <= arr[i]+arr[high] <= b and when i increases, what we should do is to decrease low and high to hold the condition. To avoid counting the same pair twice, we keep low > i, also we keep low <= high. The complexity of the following counting approach is O(N), because, in the while loop, what we can do is ++i or --low or --high and there are at most N such operations.

//count pair whose sum is in [a, b]
//arr is a sorted array with size integers.
int countPair(int arr[], int size, int a, int b) {
    int cnt = 0;
    int i = 0, low = size-1, high = size－1;
    while (i < high) {
        //find the lower bound such that arr[i] + arr[low] < a, 
        //meanwhile arr[i]+arr[low+1] >= a
         low = max(i, low);
         while (low > i && arr[i] + arr[low] >= a) --low;

        //find an upper bound such that arr[i] + arr[high] <= b 
        //meanwhile, arr[i]+arr[high+1] > b
        while (high > low && arr[i] + arr[high] > b) --high; 
        //all pairs: arr[i]+arr[low+1], arr[i]+arr[low+2],...,arr[i]+arr[high]
        //are in the rage[a, b], and we count it as follows.
        cnt += (high-low);
        ++i;
    }
    return cnt;
}