Count pairs from a given array whose sum lies from a given range - GeeksforGeeks

23 Nov, 2021
Given an array arr[] consisting of N integers and two integers L and R, the task is to count the number of pairs whose sum lies in the range [L, R].
Examples:
Input: arr[] = {5, 1, 2}, L = 4, R = 7Output: 2Explanation:The pairs satisfying the necessary conditions are as follows:
(5, 1): Sum = 5 + 1 = 6, which lies in the range [4, 7].(5, 2): Sum = 5 + 2 = 7, which lies in the range [4, 7].
(5, 1): Sum = 5 + 1 = 6, which lies in the range [4, 7].
(5, 2): Sum = 5 + 2 = 7, which lies in the range [4, 7].
Therefore, the count of such pairs is 2.
Input: arr[] = {5, 1, 2, 4, 3}, L = 5, R = 8Output: 7
Naive Approach: The simplest approach to solve the given problem is to generate all possible pairs and count those pairs whose sum lies over the range [L, R]. After checking for all the pairs, print the total count of pairs. 
Time Complexity: O(N2)Auxiliary Space: O(1)
Efficient Approach: The above approach can also be optimized by sorting the array then performing a Binary Search to find the number of elements for each array element arr[i] such that the sum will not exceed the given range. Follow the steps to solve the problem:
Sort the array arr[] in increasing order.
Initialize variables say, right as N – 1 and count as 0 to store numbers of pairs whose sum lies over the range [L, R].
Iterate until the right is greater than 0 and perform the following steps:Find the starting index of the element whose sum with arr[right] is greater than or equal to L, and store it in a variable, say start.Find the ending index of the element before arr[right] whose sum with arr[right] is less than or equal to R, and then store it in a variable, say end.If the end is greater than equal to the start, then add (end – start + 1) to the count representing the number of elements whose sum with the current element lies over the range [L, R] and then decrement right by 1.
Find the starting index of the element whose sum with arr[right] is greater than or equal to L, and store it in a variable, say start.
Find the ending index of the element before arr[right] whose sum with arr[right] is less than or equal to R, and then store it in a variable, say end.
If the end is greater than equal to the start, then add (end – start + 1) to the count representing the number of elements whose sum with the current element lies over the range [L, R] and then decrement right by 1.
After completing the above steps, print the value of count as the result.
Below is the implementation of the above approach:
C++
Java
Python3
C#
Javascript
// C++ program for the above approach #include <bits/stdc++.h>using namespace std; // Function to count pairs whose// sum lies over the range [L, R]int countPairSum(int arr[], int L,                 int R, int N){    // Sort the given array    sort(arr, arr + N);     int right = N - 1, count = 0;     // Iterate until right > 0    while (right > 0) {         // Starting index of element        // whose sum with arr[right] >= L        auto it1            = lower_bound(arr, arr + N,                          L - arr[right]);         int start = it1 - arr;         // Ending index of element        // whose sum with arr[right] <= R        auto it2            = upper_bound(arr, arr + N,                          R - arr[right]);         --it2;         int end = it2 - arr;         // Update the value of end        end = min(end, right - 1);         // Add the count of elements        // to the variable count        if (end - start >= 0) {            count += (end - start + 1);        }         right--;    }     // Return the value of count    return count;} // Driver Codeint main(){    int arr[] = { 5, 1, 2, 4, 3 };    int L = 5, R = 8;    int N = sizeof(arr) / sizeof(arr[0]);    cout << countPairSum(arr, L, R, N);     return 0;}
// Java program for the above approachimport java.util.*; class GFG{    static int lowerBound(int[] a, int low, int high, int element){        while(low < high){            int middle = low + (high - low)/2;            if(element > a[middle])                low = middle + 1;            else                high = middle;        }        return low;    }      static int upperBound(int[] a, int low, int high, int element){        while(low < high){            int middle = low + (high - low)/2;            if(a[middle] > element)                high = middle;            else                low = middle + 1;        }        return low;    }// Function to count pairs whose// sum lies over the range [L, R]static int countPairSum(int arr[], int L,                 int R, int N){    // Sort the given array    Arrays.sort(arr);     int right = N - 1, count = 0;     // Iterate until right > 0    while (right > 0) {         // Starting index of element        // whose sum with arr[right] >= L        int it1            = lowerBound(arr, 0,N,                          L - arr[right]);       it1++;        int start = it1 - arr[0];         // Ending index of element        // whose sum with arr[right] <= R        int it2            = upperBound(arr, 0,N,                          R - arr[right]);          int end = it2 - arr[0];         // Update the value of end        end = Math.min(end, right - 1);         // Add the count of elements        // to the variable count        if (end - start >= 0) {            count += (end - start +1);        }         right--;    }     // Return the value of count    return count;} // Driver Codepublic static void main(String[] args){    int arr[] = { 5, 1, 2, 4, 3 };    int L = 5, R = 8;    int N = arr.length;    System.out.print(countPairSum(arr, L, R, N)); }} // This code is contributed by gauravrajput1
# Python3 program for the above approachfrom bisect import bisect_left, bisect_right # Function to count pairs whose# sum lies over the range [L, R]def countPairSum(arr, L, R, N):         # Sort the given array    arr.sort()     right = N - 1    count = 0     # Iterate until right > 0    while (right > 0):         # Starting index of element        # whose sum with arr[right] >= L        it1 = bisect_left(arr, L - arr[right])         start = it1         # Ending index of element        # whose sum with arr[right] <= R        it2 = bisect_right(arr, R - arr[right])         it2 -= 1        end = it2         # Update the value of end        end = min(end, right - 1)         # Add the count of elements        # to the variable count        if (end - start >= 0):            count += (end - start + 1)         right -= 1     # Return the value of count    return count # Driver Codeif __name__ == '__main__':         arr = [ 5, 1, 2, 4, 3 ]    L = 5    R = 8    N = len(arr)         print(countPairSum(arr, L, R, N))     # This code is contributed by SURENDRA_GANGWAR
// C# program for the above approachusing System; public class GFG {    static int lowerBound(int[] a, int low, int high, int element) {        while (low < high) {            int middle = low + (high - low) / 2;            if (element > a[middle])                low = middle + 1;            else                high = middle;        }        return low;    }     static int upperBound(int[] a, int low, int high, int element) {        while (low < high) {            int middle = low + (high - low) / 2;            if (a[middle] > element)                high = middle;            else                low = middle + 1;        }        return low;    }     // Function to count pairs whose    // sum lies over the range [L, R]    static int countPairSum(int []arr, int L, int R, int N)    {               // Sort the given array        Array.Sort(arr);         int right = N - 1, count = 0;         // Iterate until right > 0        while (right > 0) {             // Starting index of element            // whose sum with arr[right] >= L            int it1 = lowerBound(arr, 0, N, L - arr[right]);            it1++;            int start = it1 - arr[0];             // Ending index of element            // whose sum with arr[right] <= R            int it2 = upperBound(arr, 0, N, R - arr[right]);             int end = it2 - arr[0];             // Update the value of end            end = Math.Min(end, right - 1);             // Add the count of elements            // to the variable count            if (end - start >= 0) {                count += (end - start + 1);            }             right--;        }         // Return the value of count        return count;    }     // Driver Code    public static void Main(String[] args) {        int []arr = { 5, 1, 2, 4, 3 };        int L = 5, R = 8;        int N = arr.Length;        Console.Write(countPairSum(arr, L, R, N));     }} // This code is contributed by gauravrajput1
<script>// javascript program for the above approach    function lowerBound(a , low , high , element) {        while (low < high) {            var middle = low + parseInt((high - low) / 2);            if (element > a[middle])                low = middle + 1;            else                high = middle;        }        return low;    }     function upperBound(a , low , high , element) {        while (low < high) {            var middle = low + parseInt((high - low) / 2);            if (a[middle] > element)                high = middle;            else                low = middle + 1;        }        return low;    }     // Function to count pairs whose    // sum lies over the range [L, R]    function countPairSum(arr , L , R , N)    {             // Sort the given array        arr.sort((a,b)=>a-b);         var right = N - 1, count = 0;         // Iterate until right > 0        while (right > 0) {             // Starting index of element            // whose sum with arr[right] >= L            var it1 = lowerBound(arr, 0, N, L - arr[right]);            it1++;            var start = it1 - arr[0];             // Ending index of element            // whose sum with arr[right] <= R            var it2 = upperBound(arr, 0, N, R - arr[right]);            var end = it2 - arr[0];             // Update the value of end            end = Math.min(end, right - 1);             // Add the count of elements            // to the variable count            if (end - start >= 0) {                count += (end - start + 1);            }             right--;        }         // Return the value of count        return count;    }     // Driver Code        var arr = [ 5, 1, 2, 4, 3 ];        var L = 5, R = 8;        var N = arr.length;        document.write(countPairSum(arr, L, R, N)); // This code is contributed by gauravrajput1</script>
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Time Complexity: O(N*log N)Auxiliary Space: O(1)
SURENDRA_GANGWAR
coder_nero
GauravRajput1
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