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# Make an array strictly increasing by repeatedly subtracting and adding arr[i – 1] – (i – 1) to adjacent indices

• Difficulty Level : Medium
• Last Updated : 08 Jun, 2021

Given an array arr[] consisting of N positive integers, the task is to check whether the given array arr[] can be made strictly increasing such that for any index i from the range [1, N – 1], if (arr[i – 1] – (i – 1)) is at least 0, then it is added to arr[i] and subtracted from arr[i – 1]. If it is possible to make the array strictly increasing, then print Yes. Otherwise, print No.

Examples:

Input: arr[] = {1, 5, 2, 7, 6}
Output: Yes
Explanation:
Consider the following operations:

1. Choosing the index 1, the value of arr[i – 1] – (i – 1) is 1, which is at least 0. Adding 1 to arr[1] and subtracting it from arr[0], modifies the array to {0, 6, 2, 7, 6}.
2. Choosing the index 2, the value of arr[i – 1] – (i – 1) is 5, which is at least 0. Adding 5 to arr[2] and subtracting it from arr[1], modifies the array to {0, 1, 7, 7, 6}.
3. Choosing the index 3, the value of arr[i – 1] – (i – 1) is 5, which is at least 0. Adding 6 to arr[3] and subtracting it from arr[2], modifies the array to {0, 1, 2, 12, 6}.
4. Choosing the index 4, the value of arr[i – 1] – (i – 1) is 9, which is at least 0. Adding 9 to arr[4] and subtracted from arr[3], modifies the array to {0, 1, 2, 3, 15}.

After the above operations, the array becomes strictly increasing.

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

Approach: The given problem can be solved by using the Greedy Approach. Follow the steps below to solve the problem

• Traverse the given array using variable i in range [1, N – 1] and perform the following steps:
• If arr[i – 1] is at least (i – 1), then perform the following steps:
• Store the value of arr[i] – arr[i – 1] in a variable, say P.
• Update arr[i – 1] as arr[i – 1] – P.
• Update arr[i] as arr[i] + P.
• After completing the above steps, if the array arr[] is sorted, then print “Yes”. Otherwise, print “No”

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach` `#include ``using` `namespace` `std;` `// Function to check if an array``// can be made strictly increasing``void` `check(``int` `arr[], ``int` `n)``{``    ``// Traverse the given array arr[]``    ``for` `(``int` `i = 1; i < n; i++) {` `        ``if` `(arr[i - 1] >= (i - 1)) {` `            ``// Update the value of p,``            ``// arr[i], and arr[i - 1]``            ``int` `p = arr[i - 1] - (i - 1);``            ``arr[i] += p;``            ``arr[i - 1] -= p;``        ``}``    ``}` `    ``// Traverse the given array``    ``for` `(``int` `i = 1; i < n; i++) {` `        ``// Check if the array arr[] is``        ``// strictly increasing or not``        ``if` `(arr[i] <= arr[i - 1]) {` `            ``cout << ``"No"``;``            ``return``;``        ``}``    ``}` `    ``// Otherwise, array is increasing``    ``// order, print Yes``    ``cout << ``"Yes"``;``}` `// Driver Code``int` `main()``{``    ``int` `arr[] = { 1, 5, 2, 7, 6 };``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``check(arr, N);` `    ``return` `0;``}`

## Java

 `// Java program for the above approach``import` `java.io.*;` `class` `GFG{``  ` `// Function to check if an array``// can be made strictly increasing``static` `void` `check(``int` `arr[], ``int` `n)``{``    ` `    ``// Traverse the given array arr[]``    ``for``(``int` `i = ``1``; i < n; i++)``    ``{``        ``if` `(arr[i - ``1``] >= (i - ``1``))``        ``{``            ` `            ``// Update the value of p,``            ``// arr[i], and arr[i - 1]``            ``int` `p = arr[i - ``1``] - (i - ``1``);``            ``arr[i] += p;``            ``arr[i - ``1``] -= p;``        ``}``    ``}` `    ``// Traverse the given array``    ``for``(``int` `i = ``1``; i < n; i++)``    ``{``        ` `        ``// Check if the array arr[] is``        ``// strictly increasing or not``        ``if` `(arr[i] <= arr[i - ``1``])``        ``{``            ``System.out.println(``"No"``);``            ``return``;``        ``}``    ``}` `    ``// Otherwise, array is increasing``    ``// order, print Yes``    ``System.out.println(``"Yes"``);``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``int` `arr[] = { ``1``, ``5``, ``2``, ``7``, ``6` `};``    ``int` `N = arr.length;``    ` `    ``check(arr, N);``}``}` `// This code is contributed by Dharanendra L V.`

## Python3

 `# Python3 program for the above approach` `# Function to check if an array``# can be made strictly increasing``def` `check(arr, n):``    ``# Traverse the given array arr[]``    ``for` `i ``in` `range``(``1``, n):` `        ``if` `(arr[i ``-` `1``] >``=` `(i ``-` `1``)):` `            ``# Update the value of p,``            ``# arr[i], and arr[i - 1]``            ``p ``=` `arr[i ``-` `1``] ``-` `(i ``-` `1``)``            ``arr[i] ``+``=` `p``            ``arr[i ``-` `1``] ``-``=` `p` `    ``# Traverse the given array``    ``for` `i ``in` `range``(``1``, n):``        ``# Check if the array arr[] is``        ``# strictly increasing or not``        ``if` `(arr[i] <``=` `arr[i ``-` `1``]):` `            ``print` `(``"No"``)``            ``return`  `    ``# Otherwise, array is increasing``    ``# order, prYes``    ``print` `(``"Yes"``)` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ``arr ``=` `[``1``, ``5``, ``2``, ``7``, ``6``]``    ``N ``=` `len``(arr)``    ``check(arr, N)` `# This code is contributed by mohit kumar 29.`

## C#

 `// C# program for the above approach``using` `System;``using` `System.Collections.Generic;` `class` `GFG{``  ` `// Function to check if an array``// can be made strictly increasing``static` `void` `check(``int` `[]arr, ``int` `n)``{``    ` `    ``// Traverse the given array arr[]``    ``for``(``int` `i = 1; i < n; i++)``    ``{``        ``if` `(arr[i - 1] >= (i - 1))``        ``{``            ` `            ``// Update the value of p,``            ``// arr[i], and arr[i - 1]``            ``int` `p = arr[i - 1] - (i - 1);``            ``arr[i] += p;``            ``arr[i - 1] -= p;``        ``}``    ``}` `    ``// Traverse the given array``    ``for``(``int` `i = 1; i < n; i++)``    ``{``        ` `        ``// Check if the array arr[] is``        ``// strictly increasing or not``        ``if` `(arr[i] <= arr[i - 1])``        ``{``            ``Console.Write(``"No"``);``            ``return``;``        ``}``    ``}` `    ``// Otherwise, array is increasing``    ``// order, print Yes``    ``Console.Write(``"Yes"``);``}` `// Driver Code``public` `static` `void` `Main()``{``    ``int` `[]arr = { 1, 5, 2, 7, 6 };``    ``int` `N = arr.Length;``    ``check(arr, N);``}``}` `// This code is contributed by SURENDRA_GANGWAR.`

## Javascript

 ``
Output:
`Yes`

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

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