Check whether an array can be made strictly increasing by incrementing and decrementing adjacent pairs

Last Updated : 16 Dec, 2021

Given an array arr[] of size N consisting of non-negative integers. In one move ith index element of the array is decreased by 1 and (i+1)th index is increased by 1. The task is to check if there is any possibility to make the given array strictly increasing (containing non-negative integers only) by making any number of moves.

Examples:

Input: arr[3] = [1, 2, 3]
Output: Yes
Explanation: The array is already sorted in strictly increasing order.

Input: arr[2] = [2, 0]
Output: Yes
Explanation: Consider i = 0 for the 1st move arr[0] = 2-1 = 1, arr[1] = 0 + 1 = 1. Now the array becomes, [1, 1].
In 2nd move consider i = 0. So now arr[0] = 1 – 1 = 0, arr[1] = 1 + 1 = 2.
The final array becomes, arr[2] = [0, 2] which is strictly increasing.

Input: arr[3] = [0, 1, 0]
Output: No
Explanation: This array cannot be made strictly increasing containing only non negative integers by performing any number of moves.

Approach: The problem can be solved using the following mathematical observation.

Since all the array elements are non-negative, so minimum strictly increasing order of an array of size N can be: 0, 1, 2, 3 . . . (N-1).
So the minimum sum(min_sum) of first i elements (till (i-t)th index) of any such array is min_sum = (i*(i-1))/2.
Therefore, the sum of first i elements of given array(cur_sum) must satisfy the condition cur_sum ≥ min_sum .
If the condition is not satisfied, then it is not possible to make the given array strictly increasing. Consider the following example

Illustration 1:

arr[] = 4 5 1 2 3
min_sum = 0 1 3 6 10
sum(arr) = 4 9 10 12 15

As this array satisfies the condition for every i, it is possible to convert this array to strictly increasing array

Illustration 2:

arr[] = 2 3 1 0 2
min_sum = 0 1 3 6 10
sum(arr) = 2 5 6 6 8

Here at index 4 the sum of array does not satisfy the condition of having minimum sum 10. So it is not possible to change the array into a strictly increasing one.

Follow the steps mentioned below to implement the concept:

Traverse from index = 0 to index = N – 1, and for each i check if sum till that is greater than or equal to (i*(i+1))/2.
If the condition is satisfied then the array can be made strictly increasing. Otherwise, it cannot be made strictly increasing.

Follow the below implementation for the above approach:

C++

// C++ code to check if the given array
// can be made strictly increasing
#include <bits/stdc++.h>
using namespace std;
 
// Function to check if
// the array can be made strictly increasing
void CheckStrictlyIncreasing(int arr[],
                             int N)
{
    // variable to store sum till current
    // index element
    int cur_sum = 0;
    bool possible = true;
    for (int index = 0;
         index < N; index++) {
        cur_sum += arr[index];
 
        // Sum of 0, 1, ...(i)th element
        int req_sum = (index * (index + 1)) / 2;
 
        // Check if valid or not
        if (req_sum > cur_sum) {
            possible = false;
            break;
        }
    }
 
    // If can be made strictly increasing
    if (possible)
        cout << "Yes";
    else
        cout << "No";
}
 
// Driver code
int main()
{
    int arr[3] = { 1, 2, 3 };
    int N = 3;
 
    CheckStrictlyIncreasing(arr, N);
    return 0;
}

Java

// Java code to check if the given array
// can be made strictly increasing
import java.util.*;
 
class GFG{
 
// Function to check if
// the array can be made strictly increasing
static void CheckStrictlyIncreasing(int arr[],
                             int N)
{
    // variable to store sum till current
    // index element
    int cur_sum = 0;
    boolean possible = true;
    for (int index = 0;
         index < N; index++) {
        cur_sum += arr[index];
 
        // Sum of 0, 1, ...(i)th element
        int req_sum = (index * (index + 1)) / 2;
 
        // Check if valid or not
        if (req_sum > cur_sum) {
            possible = false;
            break;
        }
    }
 
    // If can be made strictly increasing
    if (possible)
        System.out.print("Yes");
    else
        System.out.print("No");
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 1, 2, 3 };
    int N = 3;
 
    CheckStrictlyIncreasing(arr, N);
}
}
 
// This code is contributed by shikhasingrajput

Python3

# Python 3 code to check if the given array
# can be made strictly increasing
 
# Function to check if
# the array can be made strictly increasing
def CheckStrictlyIncreasing(arr,
                            N):
 
    # variable to store sum till current
    # index element
    cur_sum = 0
    possible = True
    for index in range(N):
        cur_sum += arr[index]
 
        # Sum of 0, 1, ...(i)th element
        req_sum = (index * (index + 1)) // 2
 
        # Check if valid or not
        if (req_sum > cur_sum):
            possible = False
            break
 
    # If can be made strictly increasing
    if (possible):
        print("Yes")
 
    else:
        print("No")
 
# Driver code
if __name__ == "__main__":
 
    arr = [1, 2, 3]
    N = 3
 
    CheckStrictlyIncreasing(arr, N)
 
    # This code is contributed by ukasp.

C#

// C# code to check if the given array
// can be made strictly increasing
using System;
 
class GFG{
 
// Function to check if the array can
// be made strictly increasing
static void CheckStrictlyIncreasing(int []arr,
                                    int N)
{
     
    // Variable to store sum till current
    // index element
    int cur_sum = 0;
    bool possible = true;
    for(int index = 0;
            index < N; index++) 
    {
        cur_sum += arr[index];
 
        // Sum of 0, 1, ...(i)th element
        int req_sum = (index * (index + 1)) / 2;
 
        // Check if valid or not
        if (req_sum > cur_sum)
        {
            possible = false;
            break;
        }
    }
 
    // If can be made strictly increasing
    if (possible)
        Console.Write("Yes");
    else
        Console.Write("No");
}
 
// Driver code
public static void Main(String[] args)
{
    int []arr = { 1, 2, 3 };
    int N = 3;
 
    CheckStrictlyIncreasing(arr, N);
}
}
 
// This code is contributed by shikhasingrajput

Javascript

<script>
       // JavaScript code for the above approach
 
       // Function to check if
       // the array can be made strictly increasing
       function CheckStrictlyIncreasing(arr, N)
       {
            
           // variable to store sum till current
           // index element
           let cur_sum = 0;
           let possible = true;
           for (let index = 0;
               index < N; index++) {
               cur_sum += arr[index];
 
               // Sum of 0, 1, ...(i)th element
               let req_sum = (index * (index + 1)) / 2;
 
               // Check if valid or not
               if (req_sum > cur_sum) {
                   possible = false;
                   break;
               }
           }
 
           // If can be made strictly increasing
           if (possible)
               document.write("Yes");
           else
               document.write("No");
       }
 
       // Driver code
       let arr = [1, 2, 3];
       let N = 3;
 
       CheckStrictlyIncreasing(arr, N);
 
 // This code is contributed by Potta Lokesh
   </script>