Given an array a[] consisting of N integers, the task is to find the maximum possible sum that can be achieved by deducting any value, say X, from all elements of a subarray.
Examples:
Input: N = 3, a[] = {80, 48, 82}
Output: 144
Explanation:
48 can be deducted from each array element. Therefore, sum obtained = 48 * 3 = 144
Input: N = a[] = {8, 40, 77}
Output: 80
Explanation:
Subtracting 8 from all array elements generates sum 24.
Subtracting 40 from a[1] and a[2] generates sum 80.
Subtracting 77 from a[2] generates sum 77.
Therefore, maximum possible sum is 80.
Approach:
Follow the steps below to solve the problem:
- Traverse the array
- For every element, find the element which is nearest smaller on its left and nearest smaller on its right.
- Calculate the sum possible by that element calculating current_element * ( j – i – 1 ) where j and i are the indices of the nearest smaller numbers on the left and right respectively.
- Find the maximum possible sum among all of them.
Below is the implementation of the above approach:
C++
// C++ program to implement// the above approach#include <bits/stdc++.h>using namespace std;// Function to generate previous smaller // element for each array elementvector<int> findPrevious(vector<int> a, int n){ vector<int> ps(n); // The first element has no // previous smaller ps[0] = -1; // Stack to keep track of elements // that have occurred previously stack<int> Stack; // Push the first index Stack.push(0); for(int i = 1; i < n; i++) { // Pop all the elements until the previous // element is smaller than current element while (Stack.size() > 0 && a[Stack.top()] >= a[i]) Stack.pop(); // Store the previous smaller element ps[i] = Stack.size() > 0 ? Stack.top() : -1; // Push the index of the current element Stack.push(i); } // Return the array return ps;}// Function to generate next smaller element// for each array elementvector<int> findNext(vector<int> a, int n){ vector<int> ns(n); ns[n - 1] = n; // Stack to keep track of elements // that have occurring next stack<int> Stack; Stack.push(n - 1); // Iterate in reverse order // for calculating next smaller for(int i = n - 2; i >= 0; i--) { // Pop all the elements until the // next element is smaller // than current element while (Stack.size() > 0 && a[Stack.top()] >= a[i]) Stack.pop(); // Store the next smaller element ns[i] = Stack.size() > 0 ? Stack.top() : n; // Push the index of the current element Stack.push(i); } // Return the array return ns;}// Function to find the maximum sum by// subtracting same value from all// elements of a Subarrayint findMaximumSum(vector<int> a, int n){ // Stores previous smaller element vector<int> prev_smaller = findPrevious(a, n); // Stores next smaller element vector<int> next_smaller = findNext(a, n); int max_value = 0; for(int i = 0; i < n; i++) { // Calculate contribution // of each element max_value = max(max_value, a[i] * (next_smaller[i] - prev_smaller[i] - 1)); } // Return answer return max_value;}// Driver Code int main(){ int n = 3; vector<int> a{ 80, 48, 82 }; cout << findMaximumSum(a, n); return 0;}// This code is contributed by divyeshrabadiya07 |
Java
// Java Program to implement// the above approachimport java.util.*;public class GFG { // Function to find the maximum sum by // subtracting same value from all // elements of a Subarray public static int findMaximumSum(int[] a, int n) { // Stores previous smaller element int prev_smaller[] = findPrevious(a, n); // Stores next smaller element int next_smaller[] = findNext(a, n); int max_value = 0; for (int i = 0; i < n; i++) { // Calculate contribution // of each element max_value = Math.max(max_value, a[i] * (next_smaller[i] - prev_smaller[i] - 1)); } // Return answer return max_value; } // Function to generate previous smaller element // for each array element public static int[] findPrevious(int[] a, int n) { int ps[] = new int[n]; // The first element has no // previous smaller ps[0] = -1; // Stack to keep track of elements // that have occurred previously Stack<Integer> stack = new Stack<>(); // Push the first index stack.push(0); for (int i = 1; i < a.length; i++) { // Pop all the elements until the previous // element is smaller than current element while (stack.size() > 0 && a[stack.peek()] >= a[i]) stack.pop(); // Store the previous smaller element ps[i] = stack.size() > 0 ? stack.peek() : -1; // Push the index of the current element stack.push(i); } // Return the array return ps; } // Function to generate next smaller element // for each array element public static int[] findNext(int[] a, int n) { int ns[] = new int[n]; ns[n - 1] = n; // Stack to keep track of elements // that have occurring next Stack<Integer> stack = new Stack<>(); stack.push(n - 1); // Iterate in reverse order // for calculating next smaller for (int i = n - 2; i >= 0; i--) { // Pop all the elements until the // next element is smaller // than current element while (stack.size() > 0 && a[stack.peek()] >= a[i]) stack.pop(); // Store the next smaller element ns[i] = stack.size() > 0 ? stack.peek() : a.length; // Push the index of the current element stack.push(i); } // Return the array return ns; } // Driver Code public static void main(String args[]) { int n = 3; int a[] = { 80, 48, 82 }; System.out.println(findMaximumSum(a, n)); }} |
Python3
# Python3 program to implement # the above approach # Function to find the maximum sum by # subtracting same value from all # elements of a Subarray def findMaximumSum(a, n): # Stores previous smaller element prev_smaller = findPrevious(a, n) # Stores next smaller element next_smaller = findNext(a, n) max_value = 0 for i in range(n): # Calculate contribution # of each element max_value = max(max_value, a[i] * (next_smaller[i] - prev_smaller[i] - 1)) # Return answer return max_value# Function to generate previous smaller # element for each array element def findPrevious(a, n): ps = [0] * n # The first element has no # previous smaller ps[0] = -1 # Stack to keep track of elements # that have occurred previously stack = [] # Push the first index stack.append(0) for i in range(1, n): # Pop all the elements until the previous # element is smaller than current element while len(stack) > 0 and a[stack[-1]] >= a[i]: stack.pop() # Store the previous smaller element ps[i] = stack[-1] if len(stack) > 0 else -1 # Push the index of the current element stack.append(i) # Return the array return ps# Function to generate next smaller # element for each array element def findNext(a, n): ns = [0] * n ns[n - 1] = n # Stack to keep track of elements # that have occurring next stack = [] stack.append(n - 1) # Iterate in reverse order # for calculating next smaller for i in range(n - 2, -1, -1): # Pop all the elements until the # next element is smaller # than current element while (len(stack) > 0 and a[stack[-1]] >= a[i]): stack.pop() # Store the next smaller element ns[i] = stack[-1] if len(stack) > 0 else n # Push the index of the current element stack.append(i) # Return the array return ns# Driver coden = 3a = [ 80, 48, 82 ]print(findMaximumSum(a, n))# This code is contributed by Stuti Pathak |
C#
// C# Program to implement// the above approachusing System;using System.Collections.Generic;class GFG{// Function to find the maximum sum by// subtracting same value from all// elements of a Subarraypublic static int findMaximumSum(int[] a, int n){ // Stores previous smaller element int []prev_smaller = findPrevious(a, n); // Stores next smaller element int []next_smaller = findNext(a, n); int max_value = 0; for (int i = 0; i < n; i++) { // Calculate contribution // of each element max_value = Math.Max(max_value, a[i] * (next_smaller[i] - prev_smaller[i] - 1)); } // Return answer return max_value;}// Function to generate previous smaller element// for each array elementpublic static int[] findPrevious(int[] a, int n){ int []ps = new int[n]; // The first element has no // previous smaller ps[0] = -1; // Stack to keep track of elements // that have occurred previously Stack<int> stack = new Stack<int>(); // Push the first index stack.Push(0); for (int i = 1; i < a.Length; i++) { // Pop all the elements until the previous // element is smaller than current element while (stack.Count > 0 && a[stack.Peek()] >= a[i]) stack.Pop(); // Store the previous smaller element ps[i] = stack.Count > 0 ? stack.Peek() : -1; // Push the index of the current element stack.Push(i); } // Return the array return ps;}// Function to generate next smaller element// for each array elementpublic static int[] findNext(int[] a, int n){ int []ns = new int[n]; ns[n - 1] = n; // Stack to keep track of elements // that have occurring next Stack<int> stack = new Stack<int>(); stack.Push(n - 1); // Iterate in reverse order // for calculating next smaller for (int i = n - 2; i >= 0; i--) { // Pop all the elements until the // next element is smaller // than current element while (stack.Count > 0 && a[stack.Peek()] >= a[i]) stack.Pop(); // Store the next smaller element ns[i] = stack.Count > 0 ? stack.Peek() : a.Length; // Push the index of the current element stack.Push(i); } // Return the array return ns;}// Driver Codepublic static void Main(String []args){ int n = 3; int []a = { 80, 48, 82 }; Console.WriteLine(findMaximumSum(a, n));}}// This code is contributed by Amit Katiyar |
Javascript
<script>// javascript Program to implement// the above approach // Function to find the maximum sum by // subtracting same value from all // elements of a Subarray function findMaximumSum(a ,n) { // Stores previous smaller element var prev_smaller = findPrevious(a, n); // Stores next smaller element var next_smaller = findNext(a, n); var max_value = 0; for (var i = 0; i < n; i++) { // Calculate contribution // of each element max_value = Math.max(max_value, a[i] * (next_smaller[i] - prev_smaller[i] - 1)); } // Return answer return max_value; } // Function to generate previous smaller element // for each array element function findPrevious(a , n) { var ps = Array(n).fill(0); // The first element has no // previous smaller ps[0] = -1; // Stack to keep track of elements // that have occurred previously let stack = Array(); // Push the first index stack.push(0); for (var i = 1; i < a.length; i++) { // Pop all the elements until the previous // element is smaller than current element while (stack.length > 0 && a[stack[stack.length-1]] >= a[i]) stack.pop(); // Store the previous smaller element ps[i] = stack.length > 0 ?stack[stack.length-1] : -1; // Push the index of the current element stack.push(i); } // Return the array return ps; } // Function to generate next smaller element // for each array element function findNext(a , n) { var ns = Array(n).fill(0); ns[n - 1] = n; // Stack to keep track of elements // that have occurring next var stack = Array(); stack.push(n - 1); // Iterate in reverse order // for calculating next smaller for (var i = n - 2; i >= 0; i--) { // Pop all the elements until the // next element is smaller // than current element while (stack.length > 0 && a[stack[stack.length-1]] >= a[i]) stack.pop(); // Store the next smaller element ns[i] = stack.length > 0 ? stack[stack.length-1] : a.length; // Push the index of the current element stack.push(i); } // Return the array return ns; } // Driver Code var n = 3; var a = [ 80, 48, 82 ]; document.write(findMaximumSum(a, n));// This code is contributed by gauravrajput1 </script> |
Output:
144
Time Complexity: O(N)
Auxiliary Space: O(N)
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