Given an array arr[] of N integers. The task is to find the maximum value of (arr[i] + arr[j] * arr[k]) among every triplet (i, j, k) such that arr[i] < arr[j] < arr[k] and i < j < k. If there doesn’t exist any such triplets then print “-1″.
Examples:
Input: arr[]={7, 9, 3, 8, 11, 10}
Output: 106
Explanation:
The valid triplets are:
1) (7, 9, 11), and value of (arr[i] + arr[j] * arr[k]) is 106.
2) (7, 9, 10), and value of (arr[i] + arr[j] * arr[k]) is 97.
3) (7, 8, 10), and value of (arr[i] + arr[j] * arr[k]) is 87.
4) (7, 8, 11), and value of (arr[i] + arr[j] * arr[k]) is 105.
5) (3, 8, 10), and value of (arr[i] + arr[j] * arr[k]) is 83.
6) (3, 8, 11), and value of (arr[i] + arr[j] * arr[k]) is 91.
Therefore, the maximum among the values is 106Input: arr[]={1, 2, 3}
Output: 7
Naive Approach: The idea is to generate all possible valid triplets (i, j, k) and print the maximum value of arr[i] + arr[j]*arr[k] among all the triplets. Below are the steps:
- Iterate over the array using three nested loops.
- For each valid triplets check if arr[i] < arr[j] < arr[k]. If so then the triplet is valid.
- Find the value of arr[i] + arr[j]*arr[k] for all such triplets if the above condition is true and store it in the variable called value.
- Keep updating the value of above expression to maximum value among all possible triplets.
- If no valid triplet found print -1 Otherwise print the maximum value.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function that generate all valid// triplets and calculate the value// of the valid tripletsvoid max_valid_triplet(int A[], int n){ int ans = -1; // Generate all triplets for(int i = 0; i < n - 2; i++) { for(int j = i + 1; j < n - 1; j++) { for(int k = j + 1; k < n; k++) { // Check whether the triplet // is valid or not if (A[i] < A[j] && A[j] < A[k]) { int value = A[i] + A[j] * A[k]; // Update the value if (value > ans) { ans = value; } } } } } // Print the maximum value cout << (ans);}// Driver Codeint main(){ // Given array arr[] int arr[] = { 7, 9, 3, 8, 11, 10 }; int n = sizeof(arr) / sizeof(arr[0]); // Function call max_valid_triplet(arr, n); return 0;}// This code is contributed by chitranayal |
Java
// Java program for the above approachimport java.util.Scanner;class GFG { // Function that generate all valid // triplets and calculate the value // of the valid triplets static void max_valid_triplet(int A[], int n) { int ans = -1; // Generate all triplets for (int i = 0; i < n - 2; i++) { for (int j = i + 1; j < n - 1; j++) { for (int k = j + 1; k < n; k++) { // Check whether the triplet // is valid or not if (A[i] < A[j] && A[j] < A[k]) { int value = A[i] + A[j] * A[k]; // Update the value if (value > ans) { ans = value; } } } } } // Print the maximum value System.out.println(ans); } // Driver Code public static void main(String args[]) { // Given array arr[] int[] arr = new int[] { 7, 9, 3, 8, 11, 10 }; int n = arr.length; // Function Call max_valid_triplet(arr, n); }} |
Python3
# Python3 program for the above approach# Function that generate all valid# triplets and calculate the value# of the valid tripletsdef max_valid_triplet(A, n): ans = -1; # Generate all triplets for i in range(0, n - 2): for j in range(i + 1, n - 1): for k in range(j + 1, n): # Check whether the triplet # is valid or not if (A[i] < A[j] and A[j] < A[k]): value = A[i] + A[j] * A[k]; # Update the value if (value > ans): ans = value; # Print the maximum value print(ans);# Driver Codeif __name__ == '__main__': # Given array arr arr = [ 7, 9, 3, 8, 11, 10 ]; n = len(arr); # Function call max_valid_triplet(arr, n);# This code is contributed by Amit Katiyar |
C#
// C# program for the above approachusing System;class GFG{ // Function that generate all valid // triplets and calculate the value // of the valid triplets static void max_valid_triplet(int[] A, int n) { int ans = -1; // Generate all triplets for (int i = 0; i < n - 2; i++) { for (int j = i + 1; j < n - 1; j++) { for (int k = j + 1; k < n; k++) { // Check whether the triplet // is valid or not if (A[i] < A[j] && A[j] < A[k]) { int value = A[i] + A[j] * A[k]; // Update the value if (value > ans) { ans = value; } } } } } // Print the maximum value Console.WriteLine(ans); } // Driver Code public static void Main(String[] args) { // Given array []arr int[] arr = new int[] { 7, 9, 3, 8, 11, 10 }; int n = arr.Length; // Function Call max_valid_triplet(arr, n); }}// This code is contributed by gauravrajput1 |
Javascript
<script>// JavaScript program for the above approach// Function that generate all valid// triplets and calculate the value// of the valid tripletsfunction max_valid_triplet(A, n){ let ans = -1; // Generate all triplets for(let i = 0; i < n - 2; i++) { for(let j = i + 1; j < n - 1; j++) { for(let k = j + 1; k < n; k++) { // Check whether the triplet // is valid or not if (A[i] < A[j] && A[j] < A[k]) { let value = A[i] + A[j] * A[k]; // Update the value if (value > ans) { ans = value; } } } } } // Print the maximum value document.write(ans);}// Driver Code // Given array arr[] let arr = [ 7, 9, 3, 8, 11, 10 ]; let n = arr.length; // Function call max_valid_triplet(arr, n);// This code is contributed by Surbhi Tyagi.</script> |
Output:
106
Time Complexity: O(N3)
Auxiliary Space: O(1)
Efficient approach: The above method can be optimized by using TreeSet in Java. Below are the steps:
- Create two arrays. One array (left) to store the maximum element on the left side which strictly less than the present element in the original array and another array (right) to store the right side maximum of the present element in the original array as shown in the below image for array arr[] = {7, 9, 3, 8, 11, 10}:
- For the construction of the left array, we use TreeSet in Java, insert the elements into the TreeSet, use the lower() method in TreeSet which will return the greatest element in this set which is strictly less than the given element. If no such element exists in this TreeSet collection then this method returns a NULL.
- The elements in the left array will be arr[i] of the valid triplets and the elements in the right array will be arr[k] of the valid triplet.
- Now, traverse the original array from 1 to N – 1, to select arr[j] for the valid triplet.
- If left[i]!=-1 && right[i]!=-1 then there is a chance for forming triplet.
- Find the value arr[i] + arr[j]*arr[k] for all such valid triplets and update the ans according to the maximum value.
- Print the maximum value if it exists otherwise print “-1”.
Below is the implementation of the above approach:
C++
#include <iostream>#include <set>using namespace std;// Function that finds the maximum// valid tripletsint max_valid_triplet(int A[], int n){ int ans = -1; // Declare the left[] and // right[] array int left[n]; int right[n]; // Consider last element as maximum int mx = A[n - 1]; // Iterate array from the last for (int i = n - 2; i >= 0; i--) { // If present is less the maximum // update the right[i] with // previous maximum if (mx > A[i]) right[i] = mx; // Else store -1 else right[i] = -1; // Find the maximum for // the next iteration if (mx < A[i]) mx = A[i]; } set<int> s; for (int i = 1; i < n; i++) { // Insert previous element // to the set s.insert(A[i - 1]); // Search for maximum element // which is < present element auto it = s.lower_bound(A[i]); // If result is null there is // no such element exists // so store -1 at left[i] if (it == s.begin()) left[i] = -1; // Else store the result else left[i] = *(--it); } // Traverse the original array for (int i = 1; i < n - 1; i++) { // Condition for valid triplet if (left[i] != -1 && right[i] != -1) // Find the value and update // the maximum value ans = max(ans, left[i] + A[i] * right[i]); } // Return the ans return ans;}// Driver Codeint main(){ // Given array arr[] int A[] = { 7, 9, 3, 8, 11, 10 }; int n = sizeof(A) / sizeof(A[0]); // Function Call cout << max_valid_triplet(A, n) << endl; return 0;} |
Java
// Java program for the above approachimport java.util.*;class GFG { // Function that finds the maximum // valid triplets static int max_valid_triplet(int A[], int n) { int ans = -1; // Declare the left[] and // right[] array int left[] = new int[n]; int right[] = new int[n]; // Consider last element as maximum int max = A[n - 1]; // Iterate array from the last for (int i = n - 2; i >= 0; i--) { // If present is less the maximum // update the right[i] with // previous maximum if (max > A[i]) right[i] = max; // Else store -1 else right[i] = -1; // Find the maximum for // the next iteration if (max < A[i]) max = A[i]; } TreeSet<Integer> set = new TreeSet<Integer>(); for (int i = 1; i < n; i++) { // Insert previous element // to the set set.add(A[i - 1]); Integer result = set.lower(A[i]); // Search for maximum element // which is < present element // If result is null there is // no such element exists // so store -1 at left[i] if (result == null) left[i] = -1; // Else store the result else left[i] = result; } // Traverse the original array for (int i = 1; i < n - 1; i++) { // Condition for valid triplet if (left[i] != -1 && right[i] != -1) // Find the value and update // the maximum value ans = Math.max(ans, left[i] + A[i] * right[i]); } // Return the ans return ans; } // Driver Code public static void main(String args[]) { // Given array arr[] int[] A = new int[] { 7, 9, 3, 8, 11, 10 }; int n = A.length; // Function Call System.out.println(max_valid_triplet(A, n)); }} |
Python3
# Python equivalent code# Function to find the maximum valid tripletsdef max_valid_triplet(A): n = len(A) ans = -1 left = [0] * n right = [0] * n max_value = A[n - 1] # Iterate the array from the last for i in range(n - 2, -1, -1): # If the present element is less than the maximum # update the right[i] with previous maximum if max_value > A[i]: right[i] = max_value # Else store -1 else: right[i] = -1 # Find the maximum for the next iteration if max_value < A[i]: max_value = A[i] setn = set() for i in range(1, n): # Insert previous element to the set setn.add(A[i - 1]) result = None # Search for maximum element which is < present element for j in range(A[i - 1], -1, -1): if j in setn: result = j break # If result is None, there is no such element exists # so store -1 at left[i] if result is None: left[i] = -1 # Else store the result else: left[i] = result # Traverse the original array for i in range(1, n - 1): # Condition for valid triplet if left[i] != -1 and right[i] != -1: # Find the value and update the maximum value ans = max(ans, left[i] + A[i] * right[i]) # Return the ans return ans# Driver Codeif __name__ == '__main__': # Given array A = [7, 9, 3, 8, 11, 10] # Function call print(max_valid_triplet(A)) |
C#
// C# program for the above approachusing System;using System.Collections.Generic;class GFG{// Function that finds the maximum// valid tripletsstatic int max_valid_triplet(int []A, int n){ int ans = -1; // Declare the []left and // []right array int []left = new int[n]; int []right = new int[n]; // Consider last element as maximum int max = A[n - 1]; // Iterate array from the last for(int i = n - 2; i >= 0; i--) { // If present is less the maximum // update the right[i] with // previous maximum if (max > A[i]) right[i] = max; // Else store -1 else right[i] = -1; // Find the maximum for // the next iteration if (max < A[i]) max = A[i]; } SortedSet<int> set = new SortedSet<int>(); for(int i = 1; i < n; i++) { // Insert previous element // to the set set.Add(A[i - 1]); int result = set.Min; // Search for maximum element // which is < present element // If result is null there is // no such element exists // so store -1 at left[i] if (result == 0) left[i] = -1; // Else store the result else left[i] = result; } // Traverse the original array for(int i = 1; i < n - 1; i++) { // Condition for valid triplet if (left[i] != -1 && right[i] != -1) // Find the value and update // the maximum value ans = Math.Max(ans, left[i] + A[i] * right[i]); } // Return the ans return ans;}// Driver Codepublic static void Main(String []args){ // Given array []arr int[] A = new int[]{ 7, 9, 3, 8, 11, 10 }; int n = A.Length; // Function call Console.WriteLine(max_valid_triplet(A, n));}}// This code is contributed by Amit Katiyar |
Javascript
<script>// Javascript program for the above approach// Function that finds the maximum// valid tripletsfunction max_valid_triplet(A, n){ let ans = -1; // Declare the []left and // []right array let left = new Array(n); let right = new Array(n); for(let i = 0; i < n; i++) { left[i] = 0; right[i] = 0; } // Consider last element as maximum let max = A[n - 1]; // Iterate array from the last for(let i = n - 2; i >= 0; i--) { // If present is less the maximum // update the right[i] with // previous maximum if (max > A[i]) right[i] = max; // Else store -1 else right[i] = -1; // Find the maximum for // the next iteration if (max < A[i]) max = A[i]; } let set = new Set(); for(let i = 1; i < n; i++) { // Insert previous element // to the set set.add(A[i - 1]); let result = Math.min(...Array.from(set)); // Search for maximum element // which is < present element // If result is null there is // no such element exists // so store -1 at left[i] if (result == 0) left[i] = -1; // Else store the result else left[i] = result; } // Traverse the original array for(let i = 1; i < n - 1; i++) { // Condition for valid triplet if (left[i] != -1 && right[i] != -1) // Find the value and update // the maximum value ans = Math.max(ans, left[i] + A[i] * right[i]); } // Return the ans return ans;}// Driver Codelet A = [ 7, 9, 3, 8, 11, 10 ];let n = A.length;document.write(max_valid_triplet(A, n));// This code is contributed by avanitrachhadiya2155</script> |
Output:
106
Time Complexity: O(N)
Auxiliary Space: O(N)
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