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Maximize sum of array by reducing array elements to contain no triplets (i, j, k) where a[i] < a[j] and a[i] < a[k] and j <i < k
  • Difficulty Level : Expert
  • Last Updated : 22 Dec, 2020

Given an array arr[] consisting of N integers, the task is to find the maximum sum of an array formed by decreasing array elements by 1 any number of times(possibly zero) such that there are no triplets (i, j, k)( 1-based indexing ) such that arr[j] > arr[i] and arr[k] > arr[i], where 1 ≤ j < i < k ≤ N. Print the resultant array also after performing the decrement operation.

Examples: 

Input: arr[] = {1, 2, 1, 2, 1, 3, 1}
Output:
Sum = 9
Final Array = {1, 1, 1, 1, 1, 3, 1}

Input: arr[] = {2, 4, 1, 2, 3, 1, 2}
Output
Sum = 11
Final Array: {2, 4, 1, 1, 1, 1, 1} 

Naive Approach: The idea is to update the array either increasing or decreasing or first increasing and then decreasing to get the maximum sum of all the array elements after the update. Follow the below steps to solve the problem: 



  1. Traverse the given array in the range [0, N – 1].
  2. For each index j update arr[j] as min(arr[j], b[j+1]), where b[] is storing temporary array which follow the required conditions and 0 <= j < i.
  3. For each index j update arr[j] as min(arr[j], b[j-1]), where i + 1 <= j < N.
  4. Calculate the maximum sum from each generated b[].
  5. Print the array b[] having the maximum sum.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find and print maximum
// sum and corresponding array
void maximumSum(int m[], int n)
{
    int cnt = 0, ans = 0;
 
    // b stores temperory array when
    // element at index i is peak
    // c stroes final array
    vector<int> b(n), c(n);
 
    // Check the array for each index
    for (int i = 0; i < n; i++) {
 
        // Choose m[i] as peak
        b[i] = m[i];
        cnt = b[i];
 
        // Check left
        for (int j = i - 1; j >= 0; j--) {
            b[j] = min(b[j + 1], m[j]);
            cnt += b[j];
        }
 
        // Check right
        for (int j = i + 1; j < n; j++) {
            b[j] = min(b[j - 1], m[j]);
            cnt += b[j];
        }
 
        // Check if sum is maximum
        if (ans < cnt) {
            ans = cnt;
 
            // Store the current array
            for (int j = 0; j < n; j++) {
                c[j] = b[j];
            }
        }
    }
 
    // Calculate sum
    int sum = 0;
 
    for (int i = 0; i < n; i++) {
        sum += c[i];
    }
    cout << "Sum = " << sum << endl;
 
    // Print array
    cout << "Final Array = ";
    for (int i = 0; i < n; i++) {
        cout << c[i] << " ";
    }
}
 
// Drive Code
int main()
{
    // Given array
    int arr[] = { 1, 2, 1, 2, 1, 3, 1 };
 
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    maximumSum(arr, N);
 
    return 0;
}

Java




// Java program for
// the above approach
import java.util.*;
class GFG{
 
// Function to find and print maximum
// sum and corresponding array
static void maximumSum(int m[], int n)
{
  int cnt = 0, ans = 0;
 
  // b stores temperory array when
  // element at index i is peak
  // c stroes final array
  int []b = new int[n];
  int []c = new int[n];
 
  // Check the array for each index
  for (int i = 0; i < n; i++)
  {
    // Choose m[i] as peak
    b[i] = m[i];
    cnt = b[i];
 
    // Check left
    for (int j = i - 1; j >= 0; j--)
    {
      b[j] = Math.min(b[j + 1], m[j]);
      cnt += b[j];
    }
 
    // Check right
    for (int j = i + 1; j < n; j++)
    {
      b[j] = Math.min(b[j - 1], m[j]);
      cnt += b[j];
    }
 
    // Check if sum is maximum
    if (ans < cnt)
    {
      ans = cnt;
 
      // Store the current array
      for (int j = 0; j < n; j++)
      {
        c[j] = b[j];
      }
    }
  }
 
  // Calculate sum
  int sum = 0;
 
  for (int i = 0; i < n; i++)
  {
    sum += c[i];
  }
  System.out.print("Sum = "
                    sum + "\n");
 
  // Print array
  System.out.print("Final Array = ");
  for (int i = 0; i < n; i++)
  {
    System.out.print(c[i] + " ");
  }
}
 
// Drive Code
public static void main(String[] args)
{
  // Given array
  int arr[] = {1, 2, 1, 2, 1, 3, 1};
 
  int N = arr.length;
 
  // Function Call
  maximumSum(arr, N);
}
}
 
// This code is contributed by Princi Singh

Python3




# Python3 program for the above approach
 
# Function to find and print maximum
# sum and corresponding array
def maximumSum(m, n):
     
    cnt = 0
    ans = 0
     
    # b stores temperory array when
    # element at index i is peak
    # c stroes final array
    b = [0 for i in range(n)]
    c = [0 for i in range(n)]
     
    # Check the array for each index
    for i in range(n):
         
        # Choose m[i] as peak
        b[i] = m[i]
        cnt = b[i]
         
        # Check left
        for j in range(i - 1, -1, -1):
            b[j] = min(b[j + 1], m[j])
            cnt += b[j]
         
        # Check right
        for j in range(i + 1, n):
            b[j] = min(b[j - 1], m[j])
            cnt += b[j]
         
        # Check if sum is maximum
        if (ans < cnt):
            ans = cnt
             
            # Store the current array
            for j in range(n):
                c[j] = b[j]
     
    # Calculate sum and printing
    print("Sum = ", sum(c))
    print("Final Array = ", *c)
     
# Driver Code
arr = [ 1, 2, 1, 2, 1, 3, 1 ]
 
N = len(arr) // arr[0]
 
# Function call
maximumSum(arr, N)
 
# This code is contributed by dadi madhav

C#




// C# program for the above approach
using System;
 
class GFG{
 
// Function to find and print maximum
// sum and corresponding array
static void maximumSum(int []m, int n)
{
    int cnt = 0, ans = 0;
     
    // b stores temperory array when
    // element at index i is peak
    // c stroes readonly array
    int []b = new int[n];
    int []c = new int[n];
     
    // Check the array for each index
    for(int i = 0; i < n; i++)
    {
         
        // Choose m[i] as peak
        b[i] = m[i];
        cnt = b[i];
         
        // Check left
        for(int j = i - 1; j >= 0; j--)
        {
            b[j] = Math.Min(b[j + 1], m[j]);
            cnt += b[j];
        }
         
        // Check right
        for(int j = i + 1; j < n; j++)
        {
            b[j] = Math.Min(b[j - 1], m[j]);
            cnt += b[j];
        }
         
        // Check if sum is maximum
        if (ans < cnt)
        {
            ans = cnt;
             
            // Store the current array
            for(int j = 0; j < n; j++)
            {
                c[j] = b[j];
            }
        }
    }
     
    // Calculate sum
    int sum = 0;
     
    for(int i = 0; i < n; i++)
    {
        sum += c[i];
    }
    Console.Write("Sum = "
                   sum + "\n");
     
    // Print array
    Console.Write("Final Array = ");
    for(int i = 0; i < n; i++)
    {
        Console.Write(c[i] + " ");
    }
}
 
// Drive Code
public static void Main(String[] args)
{
     
    // Given array
    int []arr = { 1, 2, 1, 2, 1, 3, 1 };
     
    int N = arr.Length;
     
    // Function call
    maximumSum(arr, N);
}
}
 
// This code is contributed by Amit Katiyar
Output: 
Sum = 9
Final Array = 1 1 1 1 1 3 1

Time Complexity: O(N2), where N is the size of the given array.
Auxiliary Space: O(N)

Efficient Approach: The idea is to use a Segment Tree to solve this problem efficiently. Follow the below steps to solve the above problem:

  • Create a Segment Tree which returns the index of the smallest element in the range [l, r].
  • Search for the smallest element in [l, r]. Suppose it is at minindex.
  • Then make 2 recursive calls: 
    • Assuming that all arr[l….minindex] have been changed to arr[minindex]. Therefore, the sum of all values is given by:

 arr[minindex]*(minindex – l+1) 

  • Then make the recursive call on [minindex + 1, r] and change all arr[minindex + 1, r] to arr[minindex]. Therefore, the sum of all values is given by:

 arr[minindex]*(r – minindex + 1)

  • In the base case, if l==r, then it is the final sum of the array when l or r is peak.
  • Then simply find the optimal peak, i.e. the one with maximum value.
  • After finding the optimal peak, print the array and its sum.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
int peak, maxi = -1;
 
// For storing segment tree
int seg[4 * 500001];
 
// Initialize segment tree
void built(int l, int r, int index,
        int a[])
{
    // Base Case
    if (l == r) {
        seg[index] = l;
        return;
    }
 
    // Get mid
    int m = l + (r - l) / 2;
 
    // Recurr from l to m
    built(l, m, 2 * index, a);
 
    // Recurr from m+1 to r
    built(m + 1, r, 2 * index + 1, a);
 
    if (a[seg[2 * index]]
        < a[seg[2 * index + 1]])
        seg[index] = seg[2 * index];
    else
        seg[index] = seg[2 * index + 1];
}
 
// Query to find minimum value index
// between l and r
int query(int l, int r, int s, int e,
        int index, int a[])
{
    // If segment is invalid
    if (s > r || e < l)
        return -1;
 
    // If segment is inside the
    // desired segment
    if (s >= l && e <= r)
        return seg[index];
 
    // Find the mid
    int m = s + (e - s) / 2;
 
    // Recurr for the left
    int d1 = query(l, r, s, m,
                2 * index, a);
 
    // Recurr for the right
    int d2 = query(l, r, m + 1, e,
                2 * index + 1, a);
 
    // Update the query
    if (d1 == -1)
        return d2;
    if (d2 == -1)
        return d1;
    if (a[d1] < a[d2])
        return d1;
    else
        return d2;
}
 
// Function for finding the optimal peak
void optimalPeak(int l, int r, int value,
                int n, int a[])
{
    if (l > r)
        return;
 
    // Check if its the peak
    if (l == r) {
 
        // Update the value for the
        // maximum sum
        if (value + a[l] > maxi) {
            maxi = a[l] + value;
            peak = l;
            return;
        }
        return;
    }
 
    // Index of minimum element in
    // l and r
    int indexmini = query(l, r, 0,
                        n - 1, 1, a);
 
    int value1 = a[indexmini]
                * (indexmini - l + 1);
 
    // Recurr right of minimum index
    optimalPeak(indexmini + 1, r,
                value + value1, n, a);
 
    // Update the max and peak value
    if (indexmini + 1 > r) {
        if (value + value1 > maxi) {
            maxi = value + value1;
            peak = indexmini;
        }
    }
 
    int value2 = a[indexmini]
                * (r - indexmini + 1);
 
    // Recurr left of minimum index
    optimalPeak(l, indexmini - 1,
                value + value2, n, a);
 
    // Update the max and peak value
    if (l > indexmini - 1) {
        if (value + value2 > maxi) {
            maxi = value + value2;
            peak = l;
        }
    }
}
 
// Print maximum sum and the array
void maximumSum(int a[], int n)
{
    // Initialize segment tree
    built(0, n - 1, 1, a);
 
    // Get the peak
    optimalPeak(0, n - 1, 0, n, a);
 
    // Store the required array
    int ans[n];
    ans[peak] = a[peak];
 
    // Update the ans[]
    for (int i = peak + 1; i < n; i++) {
        ans[i] = min(ans[i - 1], a[i]);
    }
 
    for (int i = peak - 1; i >= 0; i--) {
        ans[i] = min(a[i], ans[i + 1]);
    }
 
    // Find the maximum sum
    int sum = 0;
    for (int i = 0; i < n; i++) {
        sum += ans[i];
    }
 
    // Print sum and optimal array
    cout << "Sum = "
        << sum << endl;
 
    cout << "Final Array = ";
    for (int i = 0; i < n; i++) {
        cout << ans[i] << " ";
    }
}
 
// Drive Code
int main()
{
    // Given array
    int arr[] = { 1, 2, 1, 2, 1, 3, 1 };
 
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    maximumSum(arr, N);
 
    return 0;
}

Java




// Java program for the above approach
 
 
import java.util.*;
 
class GFG{
static int peak, maxi = -1;
 
// For storing segment tree
static int []seg = new int[4 * 500001];
 
// Initialize segment tree
static void built(int l, int r, int index,
        int a[])
{
    // Base Case
    if (l == r) {
        seg[index] = l;
        return;
    }
 
    // Get mid
    int m = l + (r - l) / 2;
 
    // Recurr from l to m
    built(l, m, 2 * index, a);
 
    // Recurr from m+1 to r
    built(m + 1, r, 2 * index + 1, a);
 
    if (a[seg[2 * index]]
        < a[seg[2 * index + 1]])
        seg[index] = seg[2 * index];
    else
        seg[index] = seg[2 * index + 1];
}
 
// Query to find minimum value index
// between l and r
static int query(int l, int r, int s, int e,
        int index, int a[])
{
    // If segment is invalid
    if (s > r || e < l)
        return -1;
 
    // If segment is inside the
    // desired segment
    if (s >= l && e <= r)
        return seg[index];
 
    // Find the mid
    int m = s + (e - s) / 2;
 
    // Recurr for the left
    int d1 = query(l, r, s, m,
                2 * index, a);
 
    // Recurr for the right
    int d2 = query(l, r, m + 1, e,
                2 * index + 1, a);
 
    // Update the query
    if (d1 == -1)
        return d2;
    if (d2 == -1)
        return d1;
    if (a[d1] < a[d2])
        return d1;
    else
        return d2;
}
 
// Function for finding the optimal peak
static void optimalPeak(int l, int r, int value,
                int n, int a[])
{
    if (l > r)
        return;
 
    // Check if its the peak
    if (l == r) {
 
        // Update the value for the
        // maximum sum
        if (value + a[l] > maxi) {
            maxi = a[l] + value;
            peak = l;
            return;
        }
        return;
    }
 
    // Index of minimum element in
    // l and r
    int indexmini = query(l, r, 0,
                        n - 1, 1, a);
 
    int value1 = a[indexmini]
                * (indexmini - l + 1);
 
    // Recurr right of minimum index
    optimalPeak(indexmini + 1, r,
                value + value1, n, a);
 
    // Update the max and peak value
    if (indexmini + 1 > r) {
        if (value + value1 > maxi) {
            maxi = value + value1;
            peak = indexmini;
        }
    }
 
    int value2 = a[indexmini]
                * (r - indexmini + 1);
 
    // Recurr left of minimum index
    optimalPeak(l, indexmini - 1,
                value + value2, n, a);
 
    // Update the max and peak value
    if (l > indexmini - 1) {
        if (value + value2 > maxi) {
            maxi = value + value2;
            peak = l;
        }
    }
}
 
// Print maximum sum and the array
static void maximumSum(int a[], int n)
{
    // Initialize segment tree
    built(0, n - 1, 1, a);
 
    // Get the peak
    optimalPeak(0, n - 1, 0, n, a);
 
    // Store the required array
    int []ans = new int[n];
    ans[peak] = a[peak];
 
    // Update the ans[]
    for (int i = peak + 1; i < n; i++) {
        ans[i] = Math.min(ans[i - 1], a[i]);
    }
 
    for (int i = peak - 1; i >= 0; i--) {
        ans[i] = Math.min(a[i], ans[i + 1]);
    }
 
    // Find the maximum sum
    int sum = 0;
    for (int i = 0; i < n; i++) {
        sum += ans[i];
    }
 
    // Print sum and optimal array
    System.out.print("Sum = "
        + sum +"\n");
 
    System.out.print("Final Array = ");
    for (int i = 0; i < n; i++) {
        System.out.print(ans[i]+ " ");
    }
}
 
// Drive Code
public static void main(String[] args)
{
    // Given array
    int arr[] = { 1, 2, 1, 2, 1, 3, 1 };
 
    int N = arr.length;
 
    // Function Call
    maximumSum(arr, N);
 
}
}
 
// This code contributed by gauravrajput1

Python3




# Python3 program for the above approach
peak = 0
maxi = -1
 
# For storing segment tree
seg = [0 for i in range(4 * 500001)]
 
# Initialize segment tree
def built(l, r, index, a):
     
    # Base Case
    if (l == r):
        seg[index] = l
        return
 
    # Get mid
    m = int(l + (r - l) / 2)
 
    # Recurr from l to m
    built(l, m, 2 * index, a)
 
    # Recurr from m+1 to r
    built(m + 1, r, 2 * index + 1, a)
 
    if (a[seg[2 * index]] <
        a[seg[2 * index + 1]]):
        seg[index] = seg[2 * index]
    else:
        seg[index] = seg[2 * index + 1]
 
# Query to find minimum value index
# between l and r
def query(l, r, s, e, index, a):
 
    # If segment is invalid
    if (s > r or e < l):
        return -1
 
    # If segment is inside the
    # desired segment
    if (s >= l and e <= r):
        return seg[index]
 
    # Find the mid
    m = int(s + (e - s) / 2)
 
    # Recurr for the left
    d1 = query(l, r, s, m,
               2 * index, a)
 
    # Recurr for the right
    d2 = query(l, r, m + 1, e,
             2 * index + 1, a)
 
    # Update the query
    if (d1 == -1):
        return d2
    if (d2 == -1):
        return d1
    if (a[d1] < a[d2]):
        return d1
    else:
        return d2
 
# Function for finding the optimal peak
def optimalPeak(l, r, value, n, a):
     
    global maxi, peak
     
    if (l > r):
        return
 
    # Check if its the peak
    if (l == r):
         
        # Update the value for the
        # maximum sum
        if (value + a[l] > maxi):
            maxi = a[l] + value
            peak = l
            return
             
        return
 
    # Index of minimum element in
    # l and r
    indexmini = query(l, r, 0, n - 1, 1, a)
    value1 = a[indexmini] * (indexmini - l + 1)
     
    # Recurr right of minimum index
    optimalPeak(indexmini + 1, r,
                    value + value1, n, a)
 
    # Update the max and peak value
    if (indexmini + 1 > r):
        if (value + value1 > maxi):
            maxi = value + value1
            peak = indexmini
 
    value2 = (a[indexmini] *
            (r - indexmini + 1))
 
    # Recurr left of minimum index
    optimalPeak(l, indexmini - 1,
                value + value2, n, a)
 
    # Update the max and peak value
    if (l > indexmini - 1):
        if (value + value2 > maxi):
            maxi = value + value2
            peak = l
 
# Print maximum sum and the array
def maximumSum(a, n):
 
    # Initialize segment tree
    built(0, n - 1, 1, a)
 
    # Get the peak
    optimalPeak(0, n - 1, 0, n, a)
 
    # Store the required array
    ans = [0 for i in range(n)]
 
    ans[peak] = a[peak]
 
    # Update the ans[]
    for i in range(peak + 1, n):
        ans[i] = min(ans[i - 1], a[i])
    for i in range(peak - 1, -1, -1):
        ans[i] = min(a[i], ans[i + 1])
 
    # Find the maximum sum
    Sum = 0
    Sum = sum(ans)
 
    # Print sum and optimal array
    print("Sum = ", Sum)
 
    print("Final Array = ", end = "")
    print(*ans, sep = " ")
 
# Driver Code
 
# Given array
arr = [ 1, 2, 1, 2, 1, 3, 1 ]
N = len(arr)
 
# Function Call
maximumSum(arr, N)
 
# This code is contributed by rag2127

C#




// C# program for
// the above approach
using System;
class GFG{
 
static int peak, maxi = -1;
 
// For storing segment tree
static int []seg = new int[4 * 500001];
 
// Initialize segment tree
static void built(int l, int r,
                int index, int []a)
{
// Base Case
if (l == r)
{
    seg[index] = l;
    return;
}
 
// Get mid
int m = l + (r - l) / 2;
 
// Recurr from l to m
built(l, m, 2 * index, a);
 
// Recurr from m+1 to r
built(m + 1, r, 2 * index + 1, a);
 
if (a[seg[2 * index]] <
    a[seg[2 * index + 1]])
    seg[index] = seg[2 * index];
else
    seg[index] = seg[2 * index + 1];
}
 
// Query to find minimum value index
// between l and r
static int query(int l, int r,
                int s, int e,
                int index, int []a)
{
// If segment is invalid
if (s > r || e < l)
    return -1;
 
// If segment is inside the
// desired segment
if (s >= l && e <= r)
    return seg[index];
 
// Find the mid
int m = s + (e - s) / 2;
 
// Recurr for the left
int d1 = query(l, r, s, m,
                2 * index, a);
 
// Recurr for the right
int d2 = query(l, r, m + 1, e,
                2 * index + 1, a);
 
// Update the query
if (d1 == -1)
    return d2;
if (d2 == -1)
    return d1;
if (a[d1] < a[d2])
    return d1;
else
    return d2;
}
 
// Function for finding the optimal peak
static void optimalPeak(int l, int r,
                        int value, int n, int []a)
{
if (l > r)
    return;
 
// Check if its the peak
if (l == r)
{
    // Update the value for the
    // maximum sum
    if (value + a[l] > maxi)
    {
    maxi = a[l] + value;
    peak = l;
    return;
    }
    return;
}
 
// Index of minimum element in
// l and r
int indexmini = query(l, r, 0,
                        n - 1, 1, a);
 
int value1 = a[indexmini] *
                (indexmini - l + 1);
 
// Recurr right of minimum index
optimalPeak(indexmini + 1, r,
            value + value1, n, a);
 
// Update the max and peak value
if (indexmini + 1 > r)
{
    if (value + value1 > maxi)
    {
    maxi = value + value1;
    peak = indexmini;
    }
}
 
int value2 = a[indexmini] *
                (r - indexmini + 1);
 
// Recurr left of minimum index
optimalPeak(l, indexmini - 1,
            value + value2, n, a);
 
// Update the max and peak value
if (l > indexmini - 1)
{
    if (value + value2 > maxi)
    {
    maxi = value + value2;
    peak = l;
    }
}
}
 
// Print maximum sum and the array
static void maximumSum(int []a, int n)
{
// Initialize segment tree
built(0, n - 1, 1, a);
 
// Get the peak
optimalPeak(0, n - 1, 0, n, a);
 
// Store the required array
int []ans = new int[n];
ans[peak] = a[peak];
 
// Update the ans[]
for (int i = peak + 1; i < n; i++)
{
    ans[i] = Math.Min(ans[i - 1], a[i]);
}
 
for (int i = peak - 1; i >= 0; i--)
{
    ans[i] = Math.Min(a[i], ans[i + 1]);
}
 
// Find the maximum sum
int sum = 0;
for (int i = 0; i < n; i++)
{
    sum += ans[i];
}
 
// Print sum and optimal array
Console.Write("Sum = " +
                sum + "\n");
 
Console.Write("Final Array = ");
for (int i = 0; i < n; i++)
{
    Console.Write(ans[i] + " ");
}
}
 
// Drive Code
public static void Main(String[] args)
{
// Given array
int []arr = {1, 2, 1, 2, 1, 3, 1};
 
int N = arr.Length;
 
// Function Call
maximumSum(arr, N);
}
}
 
// This code is contributed by shikhasingrajput
Output: 
Sum = 9
Final Array = 1 1 1 1 1 3 1

Time Complexity: O(N*logN), where N is the size of the given array.
Auxiliary Space: O(N)

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