Подсчитать количество троек в массиве с суммой в диапазоне [a, b]
Учитывая массив различных целых чисел и диапазон [a, b], задача состоит в том, чтобы подсчитать количество троек, сумма которых находится в диапазоне [a, b].
Примеры:
Ввод: arr [] = {8, 3, 5, 2} диапазон = [7, 11] Выход: 1 Есть только одна тройка {2, 3, 5} имея сумму 10 в диапазоне [7, 11]. Ввод: arr [] = {2, 7, 5, 3, 8, 4, 1, 9} диапазон = [8, 16] Выход: 36
A naive approach is to run three loops to consider all the triplets one by one. Find the sum of each triplet and increment the count if the sum lies in a given range [a, b].
Below is the implementation of the above approach:
C++
// C++ program to count triplets with // sum that lies in given range [a, b]. #include <bits/stdc++.h> using namespace std; // Function to count triplets int countTriplets( int arr[], int n, int a, int b) { // Initialize result int ans = 0; // Fix the first element as A[i] for ( int i = 0; i < n - 2; i++) { // Fix the second element as A[j] for ( int j = i + 1; j < n - 1; j++) { // Now look for the third number for ( int k = j + 1; k < n; k++) if (arr[i] + arr[j] + arr[k] >= a && arr[i] + arr[j] + arr[k] <= b) ans++; } } return ans; } // Driver Code int main() { int arr[] = { 2, 7, 5, 3, 8, 4, 1, 9 }; int n = sizeof arr / sizeof arr[0]; int a = 8, b = 16; cout << countTriplets(arr, n, a, b) << endl; return 0; } |
Java
// Java program to count triplets // with sum that lies in given // range [a, b]. import java.util.*; class GFG { // Function to count triplets public static int countTriplets( int []arr, int n, int a, int b) { // Initialize result int ans = 0 ; // Fix the first // element as A[i] for ( int i = 0 ; i < n - 2 ; i++) { // Fix the second // element as A[j] for ( int j = i + 1 ; j < n - 1 ; j++) { // Now look for the // third number for ( int k = j + 1 ; k < n; k++) { if (arr[i] + arr[j] + arr[k] >= a && arr[i] + arr[j] + arr[k] <= b) {ans++;} } } } return ans; } // Driver Code public static void main(String[] args) { int [] arr = { 2 , 7 , 5 , 3 , 8 , 4 , 1 , 9 }; int n = arr.length; int a = 8 , b = 16 ; System.out.println( "" + countTriplets(arr, n, a, b)); } } // This code is contributed // by Harshit Saini |
Python3
# Python3 program to count # triplets with sum that # lies in given range [a, b]. # Function to count triplets def countTriplets(arr, n, a, b): # Initialize result ans = 0 # Fix the first # element as A[i] for i in range ( 0 , n - 2 ): # Fix the second # element as A[j] for j in range (i + 1 , n - 1 ): # Now look for # the third number for k in range (j + 1 , n): if ((arr[i] + arr[j] + arr[k] > = a) and (arr[i] + arr[j] + arr[k] < = b)): ans + = 1 return ans # Driver code if __name__ = = "__main__" : arr = [ 2 , 7 , 5 , 3 , 8 , 4 , 1 , 9 ] n = len (arr) a = 8 ; b = 16 print (countTriplets(arr, n, a, b)) # This code is contributed # by Harshit Saini |
C#
// C# program to count triplets // with sum that lies in given // range [a, b]. using System; class GFG { // Function to count triplets public static int countTriplets( int []arr, int n, int a, int b) { // Initialize result int ans = 0; // Fix the first // element as A[i] for ( int i = 0; i < n - 2; i++) { // Fix the second // element as A[j] for ( int j = i + 1; j < n - 1; j++) { // Now look for the // third number for ( int k = j + 1; k < n; k++) { if (arr[i] + arr[j] + arr[k] >= a && arr[i] + arr[j] + arr[k] <= b) {ans++;} } } } return ans; } // Driver Code public static void Main() { int [] arr = {2, 7, 5, 3, 8, 4, 1, 9}; int n = arr.Length; int a = 8, b = 16; Console.WriteLine( "" + countTriplets(arr, n, a, b)); } } // This code is contributed // by Akanksha Rai(Abby_akku) |
PHP
<?php // PHP program to count triplets with // sum that lies in given range [a, b]. // Function to count triplets function countTriplets( $arr , $n , $a , $b ) { // Initialize result $ans = 0; // Fix the first element as A[i] for ( $i = 0; $i < $n - 2; $i ++) { // Fix the second element as A[j] for ( $j = $i + 1; $j < $n - 1; $j ++) { // Now look for the third number for ( $k = $j + 1; $k < $n ; $k ++) if ( $arr [ $i ] + $arr [ $j ] + $arr [ $k ] >= $a && $arr [ $i ] + $arr [ $j ] + $arr [ $k ] <= $b ) $ans ++; } } return $ans ; } // Driver Code $arr = array ( 2, 7, 5, 3, 8, 4, 1, 9 ); $n = sizeof( $arr ); $a = 8; $b = 16; echo countTriplets( $arr , $n , $a , $b ) . "
" ; // This code is contributed // by Akanksha Rai(Abby_akku) ?> |
Javascript
<script> // Javascript program to count triplets with // sum that lies in given range [a, b]. // Function to count triplets function countTriplets( arr, n, a, b) { // Initialize result var ans = 0; // Fix the first element as A[i] for ( var i = 0; i < n - 2; i++) { // Fix the second element as A[j] for ( var j = i + 1; j < n - 1; j++) { // Now look for the third number for ( var k = j + 1; k < n; k++) if (arr[i] + arr[j] + arr[k] >= a && arr[i] + arr[j] + arr[k] <= b) ans++; } } return ans; } // Driver Code var arr = [ 2, 7, 5, 3, 8, 4, 1, 9 ]; var n = arr.length; var a = 8, b = 16; document.write( countTriplets(arr, n, a, b) ); </script> |
36
Временная сложность: O (n 3 )
Эффективное решение - сначала найти количество троек, сумма которых меньше или равна верхнему пределу b в диапазоне [a, b]. Это количество троек также будет включать тройки, сумма которых меньше нижнего предела a. Вычтите количество троек, сумма которых меньше a. Конечный результат - это количество троек, сумма которых находится в диапазоне [a, b].
Алгоритм следующий:
- Find count of triplets having a sum less than or equal to b. Let this count be x.
- Find count of triplets having a sum less than a. Let this count be y.
- Final result is x-y.
To find the count of triplets having a sum less than or equal to the given value, refer Count triplets with sum smaller than a given value
Below is the implementation of the above approach:
C++
// C++ program to count triplets with // sum that lies in given range [a, b]. #include <bits/stdc++.h> using namespace std; // Function to find count of triplets having // sum less than or equal to val. int countTripletsLessThan( int arr[], int n, int val) { // sort the input array. sort(arr, arr + n); // Initialize result int ans = 0; int j, k; // to store sum int sum; // Fix the first element for ( int i = 0; i < n - 2; i++) { // Initialize other two elements as // corner elements of subarray arr[j+1..k] j = i + 1; k = n - 1; // Use Meet in the Middle concept. while (j != k) { sum = arr[i] + arr[j] + arr[k]; // If sum of current triplet // is greater, then to reduce it // decrease k. if (sum > val) k--; // If sum is less than or equal // to given value, then add // possible triplets (k-j) to result. else { ans += (k - j); j++; } } } return ans; } // Function to return count of triplets having // sum in range [a, b]. int countTriplets( int arr[], int n, int a, int b) { // to store count of triplets. int res; // Find count of triplets having sum less // than or equal to b and subtract count // of triplets having sum less than or // equal to a-1. res = countTripletsLessThan(arr, n, b) - countTripletsLessThan(arr, n, a - 1); return res; } // Driver Code int main() { int arr[] = { 2, 7, 5, 3, 8, 4, 1, 9 }; int n = sizeof arr / sizeof arr[0]; int a = 8, b = 16; cout << countTriplets(arr, n, a, b) << endl; return 0; } |
Java
// Java program to count triplets // with sum that lies in given // range [a, b]. import java.util.*; class GFG { // Function to find count of // triplets having sum less // than or equal to val. public static int countTripletsLessThan( int []arr, int n, int val) { // sort the input array. Arrays.sort(arr); // Initialize result int ans = 0 ; int j, k; // to store sum int sum; // Fix the first element for ( int i = 0 ; i < n - 2 ; i++) { // Initialize other two elements // as corner elements of subarray // arr[j+1..k] j = i + 1 ; k = n - 1 ; // Use Meet in the // Middle concept. while (j != k) { sum = arr[i] + arr[j] + arr[k]; // If sum of current triplet // is greater, then to reduce it // decrease k. if (sum > val) РЕКОМЕНДУЕМЫЕ СТАТЬИ |