Ways to choose three points with distance between the most distant points <= L

06 May, 2021
Given a set of n distinct points x1, x2, x3... xn all lying on the X-axis and an integer L, the task is to find the number of ways of selecting three points such that the distance between the most distant points is less than or equal to LNote: Order is not important i.e the points {3, 2, 1} and {1, 2, 3} represent the same set of three points
Examples: 
Input : x = {1, 2, 3, 4} L = 3 Output : 4Explanation: Ways to select three points such that the distance between the most distant points <= L are: 1) {1, 2, 3} Here distance between farthest points = 3 – 1 = 2 <= L 2) {1, 2, 4} Here distance between farthest points = 4 – 1 = 3 <= L 3) {1, 3, 4} Here distance between farthest points = 4 – 1 = 3 <= L 4) {2, 3, 4} Here distance between farthest points = 4 – 2 = 2 <= LThus, total number of ways = 4 
Naive Approach: First of all, sort the array of points to generate triplets {a, b, c} such that a and c are the farthest points of the triplet and a < b < c, since all the points are distinct. We can generate all the possible triplets and check for the condition if the distance between the two most distant points in <= L. If it holds we count this way, else we don’t 
C++
Java
Python3
C#
PHP
Javascript
// C++ program to count ways to choose// triplets such that the distance// between the farthest points <= L#include<bits/stdc++.h>using namespace std; // Returns the number of triplets with// distance between farthest points <= Lint countTripletsLessThanL(int n, int L, int* arr){    // sort to get ordered triplets so that we can    // find the distance between farthest points    // belonging to a triplet    sort(arr, arr + n);     int ways = 0;     // generate and check for all possible    // triplets: {arr[i], arr[j], arr[k]}    for (int i = 0; i < n; i++) {        for (int j = i + 1; j < n; j++) {            for (int k = j + 1; k < n; k++) {                 // Since the array is sorted the                // farthest points will be a[i]                // and a[k];                int mostDistantDistance = arr[k] - arr[i];                if (mostDistantDistance <= L) {                    ways++;                }            }        }    }     return ways;} // Driver Codeint main(){    // set of n points on the X axis    int arr[] = { 1, 2, 3, 4 };     int n = sizeof(arr) / sizeof(arr[0]);    int L = 3;    int ans = countTripletsLessThanL(n, L, arr);    cout << "Total Number of ways = " << ans << "\n";    return 0;}
// Java program to count ways to choose// triplets such that the distance// between the farthest points <= Limport java .io.*;import java .util.Arrays;class GFG {     // Returns the number of triplets with    // distance between farthest points <= L    static int countTripletsLessThanL(int n, int L,                                        int []arr)    {                 // sort to get ordered triplets        // so that we can find the        // distance between farthest        // points belonging to a triplet        Arrays.sort(arr);             int ways = 0;             // generate and check for all possible        // triplets: {arr[i], arr[j], arr[k]}        for (int i = 0; i < n; i++) {            for (int j = i + 1; j < n; j++) {                for (int k = j + 1; k < n; k++) {                         // Since the array is sorted the                    // farthest points will be a[i]                    // and a[k];                    int mostDistantDistance =                                    arr[k] - arr[i];                    if (mostDistantDistance <= L)                    {                        ways++;                    }                }            }        }             return ways;    }     // Driver Code    static public void main (String[] args)    {                 // set of n points on the X axis        int []arr = {1, 2, 3, 4};             int n =arr.length;        int L = 3;        int ans = countTripletsLessThanL(n, L, arr);        System.out.println("Total Number of ways = "                                             + ans);    }} // This code is contributed by anuj_67.
# Python3 program to count ways to choose# triplets such that the distance# between the farthest points <= L # Returns the number of triplets with# distance between farthest points <= Ldef countTripletsLessThanL(n, L, arr):         # sort to get ordered triplets so that    # we can find the distance between    # farthest points belonging to a triplet    arr.sort()     ways = 0     # generate and check for all possible    # triplets: {arr[i], arr[j], arr[k]}    for i in range(n):        for j in range(i + 1, n):            for k in range(j + 1, n):                 # Since the array is sorted the                # farthest points will be a[i]                # and a[k];                mostDistantDistance = arr[k] - arr[i]                if (mostDistantDistance <= L):                    ways += 1     return ways # Driver Codeif __name__ == "__main__":     # set of n points on the X axis    arr = [1, 2, 3, 4 ]     n = len(arr)    L = 3    ans = countTripletsLessThanL(n, L, arr)    print ("Total Number of ways =", ans) # This code is contributed by ita_c
// C# program to count ways to choose// triplets such that the distance// between the farthest points <= Lusing System;class GFG { // Returns the number of triplets with// distance between farthest points <= Lstatic int countTripletsLessThanL(int n, int L,                                     int []arr){         // sort to get ordered triplets    // so that we can find the    // distance between farthest    // points belonging to a triplet    Array.Sort(arr);     int ways = 0;     // generate and check for all possible    // triplets: {arr[i], arr[j], arr[k]}    for (int i = 0; i < n; i++) {        for (int j = i + 1; j < n; j++) {            for (int k = j + 1; k < n; k++) {                 // Since the array is sorted the                // farthest points will be a[i]                // and a[k];                int mostDistantDistance = arr[k] - arr[i];                if (mostDistantDistance <= L)                {                    ways++;                }            }        }    }     return ways;}     // Driver Code    static public void Main ()    {                 // set of n points on the X axis        int []arr = {1, 2, 3, 4};             int n =arr.Length;        int L = 3;        int ans = countTripletsLessThanL(n, L, arr);        Console.WriteLine("Total Number of ways = " + ans);    }} // This code is contributed by anuj_67.
<?php// PHP program to count ways to choose// triplets such that the distance// between the farthest points <= L // Returns the number of triplets with// distance between farthest points <= Lfunction countTripletsLessThanL($n, $L, $arr){    // sort to get ordered triplets so that    // we can find the distance between    // farthest points belonging to a triplet    sort($arr);     $ways = 0;     // generate and check for all possible    // triplets: {arr[i], arr[j], arr[k]}    for ($i = 0; $i < $n; $i++)    {        for ($j = $i + 1; $j < $n; $j++)        {            for ($k = $j + 1; $k < $n; $k++)            {                 // Since the array is sorted the                // farthest points will be a[i]                // and a[k];                $mostDistantDistance = $arr[$k] -                                       $arr[$i];                if ($mostDistantDistance <= $L)                {                    $ways++;                }            }        }    }     return $ways;} // Driver Code // set of n points on the X axis$arr = array( 1, 2, 3, 4 ); $n = sizeof($arr);$L = 3;$ans = countTripletsLessThanL($n, $L, $arr);echo "Total Number of ways = " , $ans, "\n"; // This code is contributed by akt_mit?>
<script> // javascript program to count ways to choose// triplets such that the distance// between the farthest points <= L     // Returns the number of triplets with    // distance between farthest points <= L    function countTripletsLessThanL(n , L,  arr)    {         // sort to get ordered triplets        // so that we can find the        // distance between farthest        // points belonging to a triplet        arr.sort();         var ways = 0;         // generate and check for all possible        // triplets: {arr[i], arr[j], arr[k]}        for (i = 0; i < n; i++) {            for (j = i + 1; j < n; j++) {                for (k = j + 1; k < n; k++) {                     // Since the array is sorted the                    // farthest points will be a[i]                    // and a[k];                    var mostDistantDistance = arr[k] - arr[i];                    if (mostDistantDistance <= L) {                        ways++;                    }                }            }        }         return ways;    }     // Driver Code        // set of n points on the X axis        var arr = [ 1, 2, 3, 4 ];         var n = arr.length;        var L = 3;        var ans = countTripletsLessThanL(n, L, arr);        document.write("Total Number of ways = " + ans); // This code contributed by Rajput-Ji </script>
Output:  
Total Number of ways = 4
Time Complexity: O(n3) for generating all possible triplets.
Efficient Approach: 
This problem can be solved by using Binary search.
First of all, sort the array.
Now, for each element of the array we find the number of elements which are greater than it(by maintaining a sorted order of points) and lie in the range (xi + 1, xi + L) both inclusive (Note that here all points are distinct so we need consider the elements equal to xi itself).
Doing so we find all such points where the distance between the farthest points will always be less than or equal to L.
Now let’s say for the ith point, we have M such points which are less than or equal to xi + L, then the number of ways we can select 2 points from M such points is simply M * (M – 1) / 2
C++
Java
Python3
C#
Javascript
// C++ program to count ways to choose// triplets such that the distance between// the farthest points <= L */#include<bits/stdc++.h>using namespace std; // Returns the number of triplets with the// distance between farthest points <= Lint countTripletsLessThanL(int n, int L, int* arr){    // sort the array    sort(arr, arr + n);     int ways = 0;    for (int i = 0; i < n; i++) {         // find index of element greater than arr[i] + L        int indexGreater = upper_bound(arr, arr + n,                                         arr[i] + L) - arr;         // find Number of elements between the ith        // index and indexGreater since the Numbers        // are sorted and the elements are distinct        // from the points btw these indices represent        // points within range (a[i] + 1 and a[i] + L)        // both inclusive         int numberOfElements = indexGreater - (i + 1);         // if there are at least two elements in between        // i and indexGreater find the Number of ways        // to select two points out of these         if (numberOfElements >= 2) {            ways += (numberOfElements                        * (numberOfElements - 1) / 2);        }    }     return ways;} // Driver Codeint main(){    // set of n points on the X axis    int arr[] = { 1, 2, 3, 4 };     int n = sizeof(arr) / sizeof(arr[0]);    int L = 4;     int ans = countTripletsLessThanL(n, L, arr);     cout << "Total Number of ways = " << ans << "\n";     return 0;}
// Java program to count ways to choose// triplets such that the distance between// the farthest points <= L */import java.util.*; class GFG{ // Returns the number of triplets with the// distance between farthest points <= Lstatic int countTripletsLessThanL(int n, int L,                                  int[] arr){    // sort the array    Arrays.sort(arr);     int ways = 0;    for (int i = 0; i < n; i++)    {         // find index of element greater than arr[i] + L        int indexGreater = upper_bound(arr, 0, n,                                       arr[i] + L);         // find Number of elements between the ith        // index and indexGreater since the Numbers        // are sorted and the elements are distinct        // from the points btw these indices represent        // points within range (a[i] + 1 and a[i] + L)        // both inclusive        int numberOfElements = indexGreater - (i + 1);         // if there are at least two elements in between        // i and indexGreater find the Number of ways        // to select two points out of these         if (numberOfElements >= 2)        {            ways += (numberOfElements *                    (numberOfElements - 1) / 2);        }    }    return ways;} static int upper_bound(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;} // Driver Codepublic static void main(String[] args){    // set of n points on the X axis    int arr[] = { 1, 2, 3, 4 };     int n = arr.length;    int L = 4;     int ans = countTripletsLessThanL(n, L, arr);     System.out.println("Total Number of ways = " + ans);}} // This code is contributed by 29AjayKumar
# Python program to count ways to choose# triplets such that the distance between# the farthest points <= L '''  # Returns the number of triplets with the# distance between farthest points <= Ldef countTripletsLessThanL(n, L, arr):    # sort the array    arr = sorted(arr);     ways = 0;    for i in range(n):         # find index of element greater than arr[i] + L        indexGreater = upper_bound(arr, 0, n, arr[i] + L);         # find Number of elements between the ith        # index and indexGreater since the Numbers        # are sorted and the elements are distinct        # from the points btw these indices represent        # points within range (a[i] + 1 and a[i] + L)        # both inclusive        numberOfElements = indexGreater - (i + 1);         # if there are at least two elements in between        # i and indexGreater find the Number of ways        # to select two points out of these         if (numberOfElements >= 2):            ways += (numberOfElements * (numberOfElements - 1) / 2);     return ways;  def upper_bound(a, low, high, element):    while (low < high):        middle = int(low + (high - low) / 2);        if (a[middle] > element):            high = middle;        else:            low = middle + 1;     return low;  # Driver Codeif __name__ == '__main__':    # set of n points on the X axis    arr = [1, 2, 3, 4];     n = len(arr);    L = 4;     ans = countTripletsLessThanL(n, L, arr);     print("Total Number of ways = " , ans);     # This code is contributed by 29AjayKumar
// C# program to count ways to choose// triplets such that the distance between// the farthest points <= L */using System; class GFG{ // Returns the number of triplets with the// distance between farthest points <= Lstatic int countTripletsLessThanL(int n, int L,                                int[] arr){    // sort the array    Array.Sort(arr);     int ways = 0;    for (int i = 0; i < n; i++)    {         // find index of element greater than arr[i] + L        int indexGreater = upper_bound(arr, 0, n,                                    arr[i] + L);         // find Number of elements between the ith        // index and indexGreater since the Numbers        // are sorted and the elements are distinct        // from the points btw these indices represent        // points within range (a[i] + 1 and a[i] + L)        // both inclusive        int numberOfElements = indexGreater - (i + 1);         // if there are at least two elements in between        // i and indexGreater find the Number of ways        // to select two points out of these        if (numberOfElements >= 2)        {            ways += (numberOfElements *                    (numberOfElements - 1) / 2);        }    }    return ways;} static int upper_bound(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;} // Driver Codepublic static void Main(String[] args){    // set of n points on the X axis    int []arr = { 1, 2, 3, 4 };     int n = arr.Length;    int L = 4;     int ans = countTripletsLessThanL(n, L, arr);     Console.WriteLine("Total Number of ways = " + ans);}} // This code is contributed by PrinciRaj1992
<script> // Javascript program to count ways to choose// triplets such that the distance between// the farthest points <= L // Returns the number of triplets with the// distance between farthest points <= Lfunction countTripletsLessThanL(n, L, arr){         // Sort the array    arr.sort(function(a, b){return a - b});     let ways = 0;    for(let i = 0; i < n; i++)    {                 // Find index of element greater        // than arr[i] + L        let indexGreater = upper_bound(arr, 0, n,                                       arr[i] + L);         // Find Number of elements between the ith        // index and indexGreater since the Numbers        // are sorted and the elements are distinct        // from the points btw these indices represent        // points within range (a[i] + 1 and a[i] + L)        // both inclusive        let numberOfElements = indexGreater - (i + 1);         // If there are at least two elements in between        // i and indexGreater find the Number of ways        // to select two points out of these        if (numberOfElements >= 2)        {            ways += (numberOfElements *                    (numberOfElements - 1) / 2);        }    }    return ways;} function upper_bound(a, low, high, element){    while(low < high)    {        let middle = low +          parseInt((high - low) / 2, 10);                   if (a[middle] > element)            high = middle;        else            low = middle + 1;    }    return low;} // Driver code // Set of n points on the X axislet arr = [ 1, 2, 3, 4 ];let n = arr.length;let L = 4; let ans = countTripletsLessThanL(n, L, arr); document.write("Total Number of ways = " + ans); // This code is contributed by suresh07 </script>
Output  
Total Number of ways = 4
Time Complexity:O(NlogN) where N is the number of points. 
vt_m
jit_t
ukasp
29AjayKumar
princiraj1992
Rajput-Ji
suresh07
Arrays
Geometric
Searching
Arrays
Searching
Geometric
Writing code in comment?
Please use ide.geeksforgeeks.org,
generate link and share the link here.