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Minimum sum possible by assigning every increasing/decreasing consecutive pair with values in that order
  • Last Updated : 21 Nov, 2020

Given an array arr[] of size N, the task is to find the minimum sum of positive integers that can be assigned to each array element arr[i], such that if arr[i] > arr[i+1] or arr[i – 1], then the positive integer assigned to arr[i] must exceed the integer assigned to arr[i + 1] or arr[i – 1].

Examples:

Input: arr[] = {1, 0, 2}
Output: 5
Explanation: One possible way to assign positive integers is ans[] = {2, 1, 2} such that the following conditions are satisfied: 
 

  • arr[0] > arr[1] and ans[0] > ans[1]
  • arr[1] < arr[2] and ans[1] < ans[2]

Therefore, minimum possible sum = 2 + 1 + 2 = 5.

 Input: arr[] = {1, 2, 2}
Output: 4
Explanation: One possible way to assign positive integers is ans[] = {1, 2, 1}. Therefore, the minimum possible sum = 1 + 2 + 1 = 4.



Approach: The idea is to traverse the given array from left to right and from right to left to update the answer for each element arr[i] such that the answer for arr[i] is greater than the answers for arr[i+1] and arr[i – 1] if arr[i] is greater than arr[i + 1] and arr[i – 1].  Follow the below steps to solve the problem:

  • Initialize a vector ans of size N that stores the minimum positive integer that can be assigned to each element.
  • Initialize the vector ans with 1 as each element must be assigned some positive integer.
  • Traverse the array from left to right using variable i and if arr[i] is greater than arr[i – 1] then update ans[i] as ans[i] = ans[i – 1] + 1.
  • Now, traverse the array from right to left using variable i if arr[i] is greater than arr[i+1], then update ans[i] as ans[i] = max(ans[i], ans[i + 1] + 1).
  • Now, find the sum of positive integers present in the vector ans and print it as the minimum possible 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 print the minimum sum
// of values assigned to each element
// of the array as per given conditions
void minSum(int* arr, int n)
{
    // Initialize vectors with value 1
    vector<int> ans(n, 1);
 
    // Traverse from left to right
    for (int i = 1; i < n; i++) {
 
        // Update if ans[i] > ans[i-1]
        if (arr[i] > arr[i - 1]) {
            ans[i] = max(ans[i],
                         ans[i - 1] + 1);
        }
    }
 
    // Traverse from right to left
    for (int i = n - 2; i >= 0; i--) {
 
        // Update as ans[i] > ans[i+1]
        // if arr[i]> arr[i+1]
        if (arr[i] > arr[i + 1]) {
            ans[i] = max(ans[i],
                         ans[i + 1] + 1);
        }
    }
 
    // Find the minimum sum
    int s = 0;
    for (auto x : ans) {
        s = s + x;
    }
 
    // Print the sum
    cout << s << endl;
}
 
// Driver Code
int main()
{
    // Given array arr[]
    int arr[] = { 1, 2, 2 };
 
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    minSum(arr, N);
 
    return 0;
}

Java




// Java program for the
// above approach
import java.util.*;
class GFG{
 
// Function to print the
// minimum sum of values
// assigned to each element
// of the array as per given
// conditions
static void minSum(int[] arr,
                   int n)
{
  // Initialize vectors
  // with value 1
  int[] ans = new int[n];
  Arrays.fill(ans, 1);
 
  // Traverse from left
  // to right
  for (int i = 1; i < n; i++)
  {
    // Update if ans[i] > ans[i-1]
    if (arr[i] > arr[i - 1])
    {
      ans[i] = Math.max(ans[i],
                        ans[i - 1] + 1);
    }
  }
 
  // Traverse from right to left
  for (int i = n - 2; i >= 0; i--)
  {
    // Update as ans[i] > ans[i+1]
    // if arr[i]> arr[i+1]
    if (arr[i] > arr[i + 1])
    {
      ans[i] = Math.max(ans[i],
                        ans[i + 1] + 1);
    }
  }
 
  // Find the minimum sum
  int s = 0;
  for (int x : ans)
  {
    s = s + x;
  }
 
  // Print the sum
  System.out.print(s + "\n");
}
 
// Driver Code
public static void main(String[] args)
{
  // Given array arr[]
  int arr[] = {1, 2, 2};
 
  int N = arr.length;
 
  // Function Call
  minSum(arr, N);
}
}
 
// This code is contributed by 29AjayKumar

Python3




# Python3 program for the
# above approach
 
# Function to print the minimum
# sum of values assigned to each
# element of the array as per
# given conditions
def minSum(arr, n):
 
    # Initialize vectors with
    # value 1
    ans = [1] * (n)
 
    # Traverse from left to
    # right
    for i in range(1, n):
 
        # Update if ans[i] >
        # ans[i-1]
        if (arr[i] > arr[i - 1]):
            ans[i] = max(ans[i],
                         ans[i - 1] + 1)
 
    # Traverse from right
    # to left
    for i in range(n - 2,
                   -1, -1):
 
        # Update as ans[i] >
        # ans[i+1] if arr[i] >
        # arr[i+1]
        if (arr[i] > arr[i + 1]):
            ans[i] = max(ans[i],
                         ans[i + 1] + 1)
 
    # Find the minimum sum
    s = 0
    for x in ans:
        s = s + x
 
    # Print the sum
    print(s)
 
# Driver Code
if __name__ == "__main__":
 
    # Given array arr[]
    arr = [1, 2, 2]
 
    N = len(arr)
 
    # Function Call
    minSum(arr, N)
 
# This code is contributed by Chitranayal

C#




// C# program for the
// above approach
using System;
class GFG{
 
// Function to print the
// minimum sum of values
// assigned to each element
// of the array as per given
// conditions
static void minSum(int[] arr,
                   int n)
{
  // Initialize vectors
  // with value 1
  int[] ans = new int[n];
   
  for(int i = 0; i < n; i++)
    ans[i] = 1;
 
  // Traverse from left
  // to right
  for (int i = 1; i < n; i++)
  {
    // Update if ans[i] > ans[i-1]
    if (arr[i] > arr[i - 1])
    {
      ans[i] = Math.Max(ans[i],
                        ans[i - 1] + 1);
    }
  }
 
  // Traverse from right to left
  for (int i = n - 2; i >= 0; i--)
  {
    // Update as ans[i] > ans[i+1]
    // if arr[i]> arr[i+1]
    if (arr[i] > arr[i + 1])
    {
      ans[i] = Math.Max(ans[i],
                        ans[i + 1] + 1);
    }
  }
 
  // Find the minimum sum
  int s = 0;
  foreach (int x in ans)
  {
    s = s + x;
  }
 
  // Print the sum
  Console.Write(s + "\n");
}
 
// Driver Code
public static void Main(String[] args)
{
  // Given array []arr
  int []arr = {1, 2, 2};
 
  int N = arr.Length;
 
  // Function Call
  minSum(arr, N);
}
}
 
// This code is contributed by gauravrajput1
Output
4

Time Complexity: O(N)
Auxiliary Space: O(N)

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