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Check whether Array can be made strictly increasing by shifting 1 value to the right

  • Last Updated : 14 Feb, 2022

Given an array arr[] of N positive integers, the task is to check whether an array can be made strictly increasing by shifting 1 value to its right element any number of times (i.e, for any value of i, decrement arr[i] and increment arr[i + 1] by 1). 

Note: The integer at any index after any operation must not be negative. 

Example:

Input: arr[] = {4, 5, 4}
Output: Yes
Explanation: The array can be made strictly increasing by performing the following operations:

  • Choose i = 1 and shift value 1 from A1 to A2 . Now, arr[] = {3, 6, 4}.
  • Choose i = 2 and shift value 1 from A2 to A3 . Now, arr[] = {3, 5, 5}.
  • Choose i = 2 and shift value 1 from A2 to A3 . Now, arr[] = {3, 4, 6}.

Input: arr[] = {0, 1, 0}
Output: No

 

Approach: The given problem can be solved using a greedy approach. The idea is that for any index i, the least possible strictly increasing sequence without including the negative integers is 0, 1, 2, 3, … . Hence, for each value of i in the range [0, N), the sum of integers till arr[i] must be greater than the sum of the series 0, 1, 2, …, i-1 which is equivalent to (i * (i – 1))/2. Therefore, if the given array satisfies that condition, return “Yes“, otherwise, return “No“.

Below is the implementation of the above approach:

C++




// C++ program of the above approach
#include <iostream>
using namespace std;
 
// Function to check whether the given
// array can be made strictly increasing
// by shifting 1 value to the right
bool isPossible(int arr[], int N)
{
    // Stores the sum of arr[]
    int sum = 0;
 
    // Loop to traverse the array
    for (int i = 0; i < N; i++) {
 
        // Update sum
        sum += arr[i];
 
        // If sum is less than the
        // least required value
        if (sum < i * (i + 1) / 2)
            return false;
    }
 
    // Return possible
    return true;
}
 
// Driver Code
int main()
{
    int arr[] = { 4, 5, 4 };
    int N = sizeof(arr) / sizeof(int);
 
    if (isPossible(arr, N)) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
 
    return 0;
}

Java




// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
class GFG
{
 
  // Function to check whether the given
  // array can be made strictly increasing
  // by shifting 1 value to the right
  static Boolean isPossible(int arr[], int N)
  {
     
    // Stores the sum of arr[]
    int sum = 0;
 
    // Loop to traverse the array
    for (int i = 0; i < N; i++) {
 
      // Update sum
      sum += arr[i];
 
      // If sum is less than the
      // least required value
      if (sum < i * (i + 1) / 2)
        return false;
    }
 
    // Return possible
    return true;
  }
 
  public static void main (String[] args) {
    int arr[] = { 4, 5, 4 };
    int N = arr.length;
 
    if (isPossible(arr, N)) {
      System.out.print("Yes");
    }
    else {
      System.out.print("No");
    }
  }
}
 
// This code is contributed by hrithikgarg03188

Python3




# Python code for the above approach
 
# Function to check whether the given
# array can be made strictly increasing
# by shifting 1 value to the right
def isPossible(arr, N):
 
    # Stores the sum of arr[]
    sum = 0
 
    # Loop to traverse the array
    for i in range(N):
 
        # Update sum
        sum += arr[i]
 
        # If sum is less than the
        # least required value
        if sum < i * (i + 1) / 2:
            return 0
 
    # Return possible
    return 1
 
# Driver Code
arr = [4, 5, 4]
N = len(arr)
 
if isPossible(arr, N) == 1:
    print("Yes")
else:
    print("No")
     
# This code is contributed by Potta Lokesh

C#




// C# program for the above approach
using System;
class GFG
{
 
  // Function to check whether the given
  // array can be made strictly increasing
  // by shifting 1 value to the right
  static Boolean isPossible(int []arr, int N)
  {
 
    // Stores the sum of arr[]
    int sum = 0;
 
    // Loop to traverse the array
    for (int i = 0; i < N; i++) {
 
      // Update sum
      sum += arr[i];
 
      // If sum is less than the
      // least required value
      if (sum < i * (i + 1) / 2)
        return false;
    }
 
    // Return possible
    return true;
  }
 
  public static void Main () {
    int []arr = { 4, 5, 4 };
    int N = arr.Length;
 
    if (isPossible(arr, N)) {
      Console.Write("Yes");
    }
    else {
      Console.Write("No");
    }
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript




<script>
// Javascript program of the above approach
 
 
// Function to check whether the given
// array can be made strictly increasing
// by shifting 1 value to the right
function isPossible(arr, N) {
    // Stores the sum of arr[]
    let sum = 0;
 
    // Loop to traverse the array
    for (let i = 0; i < N; i++) {
 
        // Update sum
        sum += arr[i];
 
        // If sum is less than the
        // least required value
        if (sum < i * (i + 1) / 2)
            return false;
    }
 
    // Return possible
    return true;
}
 
// Driver Code
 
let arr = [4, 5, 4];
let N = arr.length;
 
if (isPossible(arr, N)) {
    document.write("Yes");
}
else {
    document.write("No");
}
 
// This code is contributed by gfgking.
</script>
Output
Yes

Time Complexity: O(N)
Auxiliary Space: O(1)


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