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 15As 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 8Here 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 increasingvoid 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 codeint 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 increasingimport java.util.*;class GFG{// Function to check if// the array can be made strictly increasingstatic 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 codepublic 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 increasingdef 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 codeif __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 increasingusing System;class GFG{// Function to check if the array can// be made strictly increasingstatic 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 codepublic 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> |
Yes
Time Complexity: O(N)
Space Complexity: O(1)
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