Given an array of distinct integers and a sum value. Print all triplets with sum smaller than given sum value. Expected Time Complexity is O(n2).
Examples:
Input : arr[] = {-2, 0, 1, 3} sum = 2. Output : (-2, 0, 1) (-2, 0, 3) Explanation : The two triplets have sum less than 2. Input : arr[] = {5, 1, 3, 4, 7} sum = 12. Output : (1, 3, 4) (1, 3, 5) (1, 3, 7) (1, 4, 5)
A Simple Solution is to run three loops to consider all triplets one by one. For every triplet, compare the sums and print current triplet if its sum is smaller than given sum.
C++
// A Simple C++ program to count triplets with sum // smaller than a given value #include<bits/stdc++.h> using namespace std; int printTriplets( int arr[], int n, int sum) { // 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] < sum) cout << arr[i] << ", " << arr[j] << ", " << arr[k] << endl; } } } // Driver program int main() { int arr[] = {5, 1, 3, 4, 7}; int n = sizeof arr / sizeof arr[0]; int sum = 12; printTriplets(arr, n, sum); return 0; } |
Java
// A Simple Java program to // count triplets with sum // smaller than a given value import java.io.*; class GFG { static int printTriplets( int arr[], int n, int sum) { // 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] < sum) System.out.println(arr[i] + ", " + arr[j] + ", " + arr[k]); } } return 0 ; } // Driver Code public static void main (String[] args) { int arr[] = { 5 , 1 , 3 , 4 , 7 }; int n = arr.length; int sum = 12 ; printTriplets(arr, n, sum); } } // This code is contributed // by anuj_67. |
Python3
# A Simple python 3 program to count # triplets with sum smaller than a # given value def printTriplets(arr, n, sum ): # Fix the first element as A[i] for i in range ( 0 , n - 2 , 1 ): # Fix the second element as A[j] for j in range (i + 1 , n - 1 , 1 ): # Now look for the third number for k in range (j + 1 , n, 1 ): if (arr[i] + arr[j] + arr[k] < sum ): print (arr[i], "," , arr[j], "," , arr[k]) # Driver Code if __name__ = = '__main__' : arr = [ 5 , 1 , 3 , 4 , 7 ] n = len (arr) sum = 12 printTriplets(arr, n, sum ) # This code is contributed by # Sahil_Shelangia |
C#
// A Simple C# program to // count triplets with sum // smaller than a given value using System; class GFG { static int printTriplets( int [] arr, int n, int sum) { // 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] < sum) Console.WriteLine(arr[i] + ", " + arr[j] + ", " + arr[k]); } } return 0; } // Driver Code public static void Main () { int [] arr = {5, 1, 3, 4, 7}; int n = arr.Length; int sum = 12; printTriplets(arr, n, sum); } } // This code is contributed // by Mukul Singh. |
PHP
<?php // A Simple PHP program to count triplets // with sum smaller than a given value function printTriplets(& $arr , $n , $sum ) { // 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 ] < $sum ) { echo ( $arr [ $i ]); echo ( ", " ); echo ( $arr [ $j ]); echo ( ", " ); echo ( $arr [ $k ]); echo ( "\n" ); } } } } // Driver Code $arr = array (5, 1, 3, 4, 7); $n = sizeof( $arr ); $sum = 12; printTriplets( $arr , $n , $sum ); // This code is contributed // by Shivi_Aggarwal ?> |
5, 1, 3 5, 1, 4 1, 3, 4 1, 3, 7
Time complexity of above solution is O(n3).
An Efficient Solution can print triplets in O(n2) by sorting the array first, and then using method 1 of this post in a loop.
1) Sort the input array in increasing order. 2) Initialize result as 0. 3) Run a loop from i = 0 to n-2. An iteration of this loop finds all triplets with arr[i] as first element. a) Initialize other two elements as corner elements of subarray arr[i+1..n-1], i.e., j = i+1 and k = n-1 b) Move j and k toward each other until they meet, i.e., while (j = sum), then do k-- // Else for current i and j, there are (k-j) possible // third elements that satisfy the constraint. (ii) Else print elements from j to k
Below is the implementation of above idea.
C++
// C++ program to print triplets with sum smaller // than a given value #include <bits/stdc++.h> using namespace std; int printTriplets( int arr[], int n, int sum) { // Sort input array sort(arr, arr + n); // Every iteration of loop counts triplet with // first element as arr[i]. for ( int i = 0; i < n - 2; i++) { // Initialize other two elements as corner // elements of subarray arr[j+1..k] int j = i + 1, k = n - 1; // Use Meet in the Middle concept while (j < k) { // If sum of current triplet is more or equal, // move right corner to look for smaller values if (arr[i] + arr[j] + arr[k] >= sum) k--; // Else move left corner else { // This is important. For current i and j, // there are total k-j third elements. for ( int x = j + 1; x <= k; x++) cout << arr[i] << ", " << arr[j] << ", " << arr[x] << endl; j++; } } } } // Driver program int main() { int arr[] = { 5, 1, 3, 4, 7 }; int n = sizeof arr / sizeof arr[0]; int sum = 12; printTriplets(arr, n, sum); return 0; } |
Java
// Java program to print // triplets with sum smaller // than a given value import java.util.*; import java.lang.*; import java.io.*; class GFG { static void printTriplets( int arr[], int n, int sum) { // Sort input array Arrays.sort(arr); // Every iteration of loop // counts triplet with // first element as arr[i]. for ( int i = 0 ; i < n - 2 ; i++) { // Initialize other two elements // as corner elements of subarray // arr[j+1..k] int j = i + 1 , k = n - 1 ; // Use Meet in the // Middle concept while (j < k) { // If sum of current triplet // is more or equal, move right // corner to look for smaller values if (arr[i] + arr[j] + arr[k] >= sum) k--; // Else move left corner else { // This is important. For // current i and j, there // are total k-j third elements. for ( int x = j + 1 ; x <= k; x++) System.out.println(arr[i] + ", " + arr[j] + ", " + arr[x]); j++; } } } } // Driver Code public static void main(String args[]) { int arr[] = { 5 , 1 , 3 , 4 , 7 }; int n = arr.length; int sum = 12 ; printTriplets(arr, n, sum); } } // This code is contributed // by Subhadeep |
Python3
# Python3 program to print # triplets with sum smaller # than a given value def printTriplets(arr, n, sum ): # Sort input array arr.sort() # Every iteration of loop # counts triplet with # first element as arr[i]. for i in range (n - 2 ): # Initialize other two elements # as corner elements of subarray # arr[j+1..k] (j, k) = (i + 1 , n - 1 ) # Use Meet in the # Middle concept while (j < k): # If sum of current triplet # is more or equal, move right # corner to look for smaller values if (arr[i] + arr[j] + arr[k] > = sum ): k - = 1 # Else move left corner else : # This is important. For # current i and j, there # are total k-j third elements. for x in range (j + 1 , k + 1 ): print ( str (arr[i]) + ", " + str (arr[j]) + ", " + str (arr[x])) j + = 1 # Driver code if __name__ = = "__main__" : arr = [ 5 , 1 , 3 , 4 , 7 ] n = len (arr) sum = 12 printTriplets(arr, n, sum ); # This code is contributed by rutvik_56 |
C#
// C# program to print // triplets with sum smaller // than a given value using System; class GFG { static void printTriplets( int [] arr, int n, int sum) { // Sort input array Array.Sort(arr); // Every iteration of loop // counts triplet with // first element as arr[i]. for ( int i = 0; i < n - 2; i++) { // Initialize other two elements // as corner elements of subarray // arr[j+1..k] int j = i + 1, k = n - 1; // Use Meet in the // Middle concept while (j < k) { // If sum of current triplet // is more or equal, move right // corner to look for smaller values if (arr[i] + arr[j] + arr[k] >= sum) k--; // Else move left corner else { // This is important. For // current i and j, there // are total k-j third elements. for ( int x = j + 1; x <= k; x++) Console.WriteLine(arr[i] + ", " + arr[j] + ", " + arr[x]); j++; } } } } // Driver Code public static void Main() { int [] arr = { 5, 1, 3, 4, 7 }; int n = arr.Length; int sum = 12; printTriplets(arr, n, sum); } } // This code is contributed // by Akanksha Rai |
PHP
<?php // PHP program to print triplets with // sum smaller than a given value function printTriplets( $arr , $n , $sum ) { // Sort input array sort( $arr , 0); // Every iteration of loop counts // triplet with first element as arr[i]. for ( $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 ) { // If sum of current triplet is more // or equal, move right corner to // look for smaller values if ( $arr [ $i ] + $arr [ $j ] + $arr [ $k ] >= $sum ) $k --; // Else move left corner else { // This is important. For current i and j, // there are total k-j third elements. for ( $x = $j + 1; $x <= $k ; $x ++) echo $arr [ $i ] . ", " . $arr [ $j ] . ", " . $arr [ $x ] . "\n" ; $j ++; } } } } // Driver Code $arr = array (5, 1, 3, 4, 7); $n = sizeof( $arr ); $sum = 12; printTriplets( $arr , $n , $sum ); // This code is contributed // by Akanksha Rai ?> |
1, 3, 4 1, 3, 5 1, 3, 7 1, 4, 5
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