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Minimum steps required to reduce all array elements to 1 based on given steps

  • Last Updated : 22 Nov, 2021

Given an array arr[] of size N. The task is to find the minimum steps required to reduce all array elements to 1. In each step, perform the following given operation:

  • Choose any starting index, say i, and jump to the (arr[i] + i)th index, reducing ith as well as (arr[i] + i)th index by 1, follow this process until it reaches the end of the array
  • If an element is already reduced to 1, it can’t be reduced more, it remains the same.

Examples:

Input: arr[] = {4, 2, 3, 2, 2, 2, 1, 2}, N = 8
Output: 5
Explanation: Series of operations can be performed in the following way:

  • {4, 2, 3, 2, 2, 2, 1, 2}, decrement values by 1, arr[] = {4, 2, 2, 2, 2, 1, 1, 1}
  • {4, 2, 2, 2, 2, 1, 1, 1}, decrement values by 1, arr[] = {3, 2, 2, 2, 1, 1, 1, 1}
  • {3, 2, 2, 2, 1, 1, 1, 1}, decrement values by 1, arr[] = {2, 2, 2, 1, 1, 1, 1, 1}
  • {2, 2, 2, 1, 1, 1, 1, 1}, decrement values by 1, arr[] = {1, 2, 1, 1, 1, 1, 1, 1}
  • {1, 2, 1, 1, 1, 1, 1, 1}, decrement values by 1, arr[] = {1, 1, 1, 1, 1, 1, 1, 1}

So, total steps required = 5

Input: arr[] = {1, 3, 1, 2, 2}, N = 5
Output: 2

 

Approach: The given problem can be solved by dividing the problem in 4 parts :- (0 to i-1) |  i  | (i + 1) | (i + 2 to n – 1). Follow the below steps to solve the problem:

  1. Take a vector say v, which will denote how many times an element is decreased due to visiting the previous elements.
  2. Each element in v i.e v[i] denotes the count of decrement in arr[i] due to visiting the elements from 0 to (i-1).
  3. Iterate over the array arr[], and take a variable say k to store the numbers of passes that have to add in answer due to element arr[i] after making 0 to (i-1) elements equal to 1.
  4. Inside the loop, run another loop(i.e the second loop) to compute how much current arr[i] elements effect(decrease after visiting the ith element) the (i+2) to N elements.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find minimum steps
// required to reduce all array
// elements to 1
int minSteps(int arr[], int N)
{
    // Variable to store the answer
    int steps = 0;
 
    // Vector with all elements initialized
    // with 0
    vector<long long> v(N + 1, 0);
 
    // Traverse the array
    for (int i = 0; i < N; ++i) {
        // Variable to store the numbers of
        // passes that have to add in answer
        // due to element arr[i] after making
        // 0 to (i-1) elements equal to 1
        int k = max(0ll, arr[i] - 1 - v[i]);
 
        // Increment steps by K
        steps += k;
        // Update element in v
        v[i] += k;
 
        // Loop to compute how much current element
        // effect the (i+2) to N elements
        for (int j = i + 2; j <= min(i + arr[i], N); j++) {
            v[j]++;
        }
        v[i + 1] += v[i] - arr[i] + 1;
    }
    // Return steps which is the answer
    return steps;
}
 
// Driver Code
int main()
{
    int arr[] = { 4, 2, 3, 2, 2, 2, 1, 2 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    cout << minSteps(arr, N);
 
    return 0;
}

Java




// Java code for the above approach
import java.io.*;
 
class GFG
{
   
    // Function to find minimum steps
    // required to reduce all array
    // elements to 1
    static int minSteps(int arr[], int N)
    {
       
        // Variable to store the answer
        int steps = 0;
 
        // Vector with all elements initialized
        // with 0
        int v[] = new int[N + 1];
 
        // Traverse the array
        for (int i = 0; i < N; ++i)
        {
           
            // Variable to store the numbers of
            // passes that have to add in answer
            // due to element arr[i] after making
            // 0 to (i-1) elements equal to 1
            int k = Math.max(0, arr[i] - 1 - v[i]);
 
            // Increment steps by K
            steps += k;
            // Update element in v
            v[i] += k;
 
            // Loop to compute how much current element
            // effect the (i+2) to N elements
            for (int j = i + 2;
                 j <= Math.min(i + arr[i], N); j++) {
                v[j]++;
            }
            v[i + 1] += v[i] - arr[i] + 1;
        }
        // Return steps which is the answer
        return steps;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int arr[] = { 4, 2, 3, 2, 2, 2, 1, 2 };
        int N = arr.length;
 
        System.out.println(minSteps(arr, N));
    }
}
 
// This code is contributed by Potta Lokesh

Python3




# Python program for the above approach
 
# Function to find minimum steps
# required to reduce all array
# elements to 1
def minSteps(arr, N):
   
    # Variable to store the answer
    steps = 0
 
    # Vector with all elements initialized
    # with 0
    v = [0] * (N + 1)
 
    # Traverse the array
    for i in range(N):
       
        # Variable to store the numbers of
        # passes that have to add in answer
        # due to element arr[i] after making
        # 0 to (i-1) elements equal to 1
        k = max(0, arr[i] - 1 - v[i])
 
        # Increment steps by K
        steps += k
        # Update element in v
        v[i] += k
 
        # Loop to compute how much current element
        # effect the (i+2) to N elements
        for j in range(i + 2, min(i + arr[i], N) + 1):
            v[j] += 1
        v[i + 1] += v[i] - arr[i] + 1
 
    # Return steps which is the answer
    return steps
 
# Driver Code
arr = [4, 2, 3, 2, 2, 2, 1, 2]
N = len(arr)
 
print(minSteps(arr, N))
 
 # This code is contributed by gfgking.

C#




// C# code for the above approach
using System;
class GFG
{
   
    // Function to find minimum steps
    // required to reduce all array
    // elements to 1
    static int minSteps(int[] arr, int N)
    {
       
        // Variable to store the answer
        int steps = 0;
 
        // Vector with all elements initialized
        // with 0
        int[] v = new int[N + 1];
 
        // Traverse the array
        for (int i = 0; i < N; ++i)
        {
           
            // Variable to store the numbers of
            // passes that have to add in answer
            // due to element arr[i] after making
            // 0 to (i-1) elements equal to 1
            int k = Math.Max(0, arr[i] - 1 - v[i]);
 
            // Increment steps by K
            steps += k;
           
            // Update element in v
            v[i] += k;
 
            // Loop to compute how much current element
            // effect the (i+2) to N elements
            for (int j = i + 2;
                 j <= Math.Min(i + arr[i], N); j++) {
                v[j]++;
            }
            v[i + 1] += v[i] - arr[i] + 1;
        }
       
        // Return steps which is the answer
        return steps;
    }
 
    // Driver Code
    public static void Main()
    {
        int[] arr = { 4, 2, 3, 2, 2, 2, 1, 2 };
        int N = arr.Length;
 
        Console.Write(minSteps(arr, N));
    }
}
 
// This code is contributed by gfgking

Javascript




<script>
    // JavaScript program for the above approach
 
    // Function to find minimum steps
    // required to reduce all array
    // elements to 1
    const minSteps = (arr, N) => {
        // Variable to store the answer
        let steps = 0;
 
        // Vector with all elements initialized
        // with 0
        let v = new Array(N + 1).fill(0);
 
        // Traverse the array
        for (let i = 0; i < N; ++i) {
            // Variable to store the numbers of
            // passes that have to add in answer
            // due to element arr[i] after making
            // 0 to (i-1) elements equal to 1
            let k = Math.max(0, arr[i] - 1 - v[i]);
 
            // Increment steps by K
            steps += k;
            // Update element in v
            v[i] += k;
 
            // Loop to compute how much current element
            // effect the (i+2) to N elements
            for (let j = i + 2; j <= Math.min(i + arr[i], N); j++) {
                v[j]++;
            }
            v[i + 1] += v[i] - arr[i] + 1;
        }
        // Return steps which is the answer
        return steps;
    }
 
    // Driver Code
    let arr = [4, 2, 3, 2, 2, 2, 1, 2];
    let N = arr.length
 
    document.write(minSteps(arr, N));
 
    // This code is contributed by rakeshsahni
 
</script>

 
 

Output: 
5

 

Time Complexity: O(N2)
Auxiliary Space: O(N)

 


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