Maximum index a pointer can reach in N steps by avoiding a given index B

Maximum index a pointer can reach in N steps by avoiding a given index B | Set 2

Last Updated : 07 Apr, 2021

Given two integers N and B, the task is to print the maximum index in an array that can be reached, starting from the 0^th index, in N steps without placing itself at index B at any point, where in every i^th step, pointer can move i indices to the right.

Examples:

Input: N = 4, B = 6
Output: 9
Explanation: Following sequence of moves maximizes the index that can be reached.
Step 1: Initially, pos = 0. Remain in the same position.
Step 2: Move 2 indices to the right. Therefore, current position = 0 + 2 = 2.
Step 3: Move 3 indices to the right. Therefore, current position = 2 + 3 = 5.
Step 4: Move 4 indices to the right. Therefore, current position = 5 + 4 = 9.
Input: N = 2, B = 2
Output: 3

Naive Approach: Refer to the previous post for the simplest approach to solve the problem.

Time Complexity: O(N³)
Auxiliary Space: O(1)

Efficient Approach: The most optimal idea to solve the problem is based on the following observations:

Observation:
If observed carefully, the answer is either the sequence from the arithmetic sum of steps or that of the arithmetic sum of steps – 1.
This is because, the highest possible number without considering B, is reachable by not waiting (which would give the arithmetic sum).
But if B is a part of that sequence, then waiting at 0 in the first steps ensures that the sequence does not intersect with the sequence obtained without waiting (as it is always 1 behind).
Any other sequence (i.e waiting at any other point once or more number of times) will always yield a smaller maximum reachable index.

Follow the steps below to solve the problem:

Initialize two pointers i = 0 and j = 1.
Initialize a variable, say sum, to store the sum of first N natural numbers, i.e. N * (N + 1) / 2.
Initialize a variable, say cnt = 0 and another variable, say flag = false.
Iterate until cnt is less than N.
- Increment i with j.
- Increment j.
- Increment cnt.
- If at any iteration, i is equal to B, set flag = true and break out of the loop.
If flag is false, then print sum. Otherwise, print sum – 1.

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 the maximum
// index the pointer can reach
int maximumIndex(int N, int B)
{
    // Initialize two pointers
    int i = 0, j = 1;
 
    // Stores number of steps
    int cnt = 0;
 
    // Stores sum of first N
    // natural numbers
    int sum = N * (N + 1) / 2;
 
    bool flag = false;
 
    while (cnt < N) {
 
        // Increment i with j
        i += j;
 
        // Increment j with 1
        j++;
 
        // Increment count
        cnt++;
 
        // If i points to B
        if (i == B) {
 
            // Break
            flag = true;
            break;
        }
    }
 
    // Print the pointer index
    if (!flag) {
        cout << sum;
    }
    else
        cout << sum - 1;
}
 
// Driver Code
int main()
{
    // Given value of N & B
    int N = 4, B = 6;
 
    // Function call to find maximum
    // index the pointer can reach
    maximumIndex(N, B);
 
    return 0;
}

Java

// Java program for the above approach
class GFG{
     
// Function to find the maximum
// index the pointer can reach
static void maximumIndex(int N, int B)
{
     
    // Initialize two pointers
    int i = 0, j = 1;
 
    // Stores number of steps
    int cnt = 0;
 
    // Stores sum of first N
    // natural numbers
    int sum = N * (N + 1) / 2;
 
    boolean flag = false;
 
    while (cnt < N)
    {
         
        // Increment i with j
        i += j;
 
        // Increment j with 1
        j++;
 
        // Increment count
        cnt++;
 
        // If i points to B
        if (i == B)
        {
             
            // Break
            flag = true;
            break;
        }
    }
 
    // Print the pointer index
    if (!flag == true)
    {
        System.out.print(sum);
    }
    else
    {
        System.out.print(sum - 1);
    }
}
 
// Driver Code
public static void main (String[] args)
{
     
    // Given value of N & B
    int N = 4, B = 6;
 
    // Function call to find maximum
    // index the pointer can reach
    maximumIndex(N, B);
}
}
 
// This code is contributed by AnkThon

Python3

# Python3 program for the above approach
 
# Function to find the maximum
# index the pointer can reach
def maximumIndex(N, B):
     
    # Initialize two pointers
    i, j = 0, 1
 
    # Stores number of steps
    cnt = 0
 
    # Stores sum of first N
    # natural numbers
    sum = N * (N + 1) // 2
 
    flag = False
 
    while (cnt < N):
 
        # Increment i with j
        i += j
 
        # Increment j with 1
        j += 1
 
        # Increment count
        cnt += 1
 
        # If i points to B
        if (i == B):
 
            # Break
            flag = True
            break
 
    # Print the pointer index
    if (not flag):
        print (sum)
    else:
        print(sum - 1)
 
# Driver Code
if __name__ == '__main__':
     
    # Given value of N & B
    N, B = 4, 6
 
    # Function call to find maximum
    # index the pointer can reach
    maximumIndex(N, B)
 
# This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using System;
 
class GFG{
     
// Function to find the maximum
// index the pointer can reach
static void maximumIndex(int N, int B)
{
     
    // Initialize two pointers
    int i = 0, j = 1;
 
    // Stores number of steps
    int cnt = 0;
 
    // Stores sum of first N
    // natural numbers
    int sum = N * (N + 1) / 2;
 
    bool flag = false;
 
    while (cnt < N)
    {
         
        // Increment i with j
        i += j;
 
        // Increment j with 1
        j++;
 
        // Increment count
        cnt++;
 
        // If i points to B
        if (i == B)
        {
             
            // Break
            flag = true;
            break;
        }
    }
 
    // Print the pointer index
    if (!flag == true)
    {
        Console.Write(sum);
    }
    else
    {
       Console.Write(sum - 1);
    }
}
 
// Driver Code
static public void Main ()
{
     
    // Given value of N & B
    int N = 4, B = 6;
 
    // Function call to find maximum
    // index the pointer can reach
    maximumIndex(N, B);
}
}
 
// This code is contributed by avijitmondal1998

Javascript

<script>
// JavaScript program for the above approach
 
// Function to find the maximum
// index the pointer can reach
function maximumIndex(N, B)
{
    // Initialize two pointers
    let i = 0, j = 1;
 
    // Stores number of steps
    let cnt = 0;
 
    // Stores sum of first N
    // natural numbers
    let sum = Math.floor(N * (N + 1) / 2);
 
    let flag = false;
 
    while (cnt < N) {
 
        // Increment i with j
        i += j;
 
        // Increment j with 1
        j++;
 
        // Increment count
        cnt++;
 
        // If i points to B
        if (i == B) {
 
            // Break
            flag = true;
            break;
        }
    }
 
    // Print the pointer index
    if (!flag) {
        document.write(sum);
    }
    else
        document.write(sum - 1);
}
 
// Driver Code
 
    // Given value of N & B
    let N = 4, B = 6;
 
    // Function call to find maximum
    // index the pointer can reach
    maximumIndex(N, B);
 
// This code is contributed by Surbhi Tyagi.
</script>

Output:

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

Attention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the Essential Maths for CP Course at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more, please refer Complete Interview Preparation Course.

My Personal Notes

Related Articles

C++

Java

Python3

C#

Javascript