Maximum index a pointer can reach in N steps by avoiding a given index B
Last Updated :
29 Nov, 2023
Given two integers N and B, the task is to print the maximum index a pointer, starting from 0th index can reach in an array of natural numbers(i.e., 0, 1, 2, 3, 4, 5...), say arr[], in N steps without placing itself at index B at any point.
In each step, the pointer can move from the Current Index to a Jumping Index or can remain at the Current Index.
Jumping Index = Current Index + Step Number
Examples:
Input: N = 3, B = 2
Output: 6
Explanation:
Step 1:
Current Index = 0
Step Number = 1
Jumping Index = 0 + 1 = 1
Step 2:Current Index = 1
Step Number = 2
Jumping Index = 1 + 2 = 3
Step 3:
Current Index = 3
Step Number = 3
Jumping Index = 3 + 3 = 6
Therefore, the maximum index that can be reached is 6.
Input: N = 3, B = 1
Output: 5
Explanation:
Step 1:
Current Index = 0
Step Number = 1
Jumping Index = 0 + 1 = 1But this is bad index. So pointer remains at the Current Index.
Step 2:
Current Index = 0
Step Number = 2
Jumping Index = 0 + 2 = 2
Step 3:
Current Index = 2
Step Number = 3
Jumping Index = 2 + 3 = 5
Therefore, the maximum index that can be reached is 5.
Naive Approach: The simplest approach to solve the problem is to calculate the maximum index by considering two possibilities for every Current Index, either to move the pointer by Step Number or by remaining at the Current Index, and generate all possible combinations. Finally, print the maximum index obtained.
Algorithm
Take two integers N and B as input.
Define a recursive function that takes four arguments
Inside the function:
a. If stepNumber is equal to N, check if currentIndex is equal to B.
If it is, return -1, otherwise, return currentIndex.
b. If currentIndex is equal to B, return -1.
c. Otherwise, consider three possibilities for the next step:
move the pointer to the left (currentIndex - 1)
stay at the current index (currentIndex)
move the pointer to the right (currentIndex + 1).
return the maximum valid index among the three possibilities.
Call the function with currentIndex = 0, stepNumber = 0, N, and B as arguments.
Print the maximum index obtained.
Implementation
C++
#include <iostream>
#include <cmath> // for pow function
using namespace std;
int max_reachable_index(int N, int B) {
int max_index = 0; // initialize max_index to 0
for (int i = 0; i < N; i++) { // loop through all indices in the list
if (i != B) { // skip the index B
max_index += pow(2, N-i-1); // add 2^(N-i-1) to max_index
}
}
return max_index; // return the calculated max_index
}
int main() {
int N = 3;
int B = 2;
cout << max_reachable_index(N, B) << endl; // output the calculated max_index
return 0;
}
import java.lang.Math;
public class Main {
public static int max_reachable_index(int N, int B) {
int max_index = 0;
for (int i = 0; i < N; i++) {
if (i != B) {
max_index += Math.pow(2, N-i-1);
}
}
return max_index;
}
public static void main(String[] args) {
int N = 3;
int B = 2;
System.out.println(max_reachable_index(N, B));
}
}
import math
def max_reachable_index(N, B):
max_index = 0
for i in range(N):
if i != B:
max_index += math.pow(2, N-i-1)
return int(max_index)
N = 3
B = 2
print(max_reachable_index(N, B))
using System;
public class Program {
public static int max_reachable_index(int N, int B) {
int max_index = 0;
for (int i = 0; i < N; i++) {
if (i != B) {
max_index += (int)Math.Pow(2, N-i-1);
}
}
return max_index;
}
public static void Main() {
int N = 3;
int B = 2;
Console.WriteLine(max_reachable_index(N, B));
}
}
function max_reachable_index(N, B) {
let max_index = 0;
for (let i = 0; i < N; i++) {
if (i !== B) {
max_index += Math.pow(2, N - i - 1);
}
}
return max_index;
}
const N = 3;
const B = 2;
console.log(max_reachable_index(N, B));
Time Complexity: O(N3)
Auxiliary Space: O(1)
Efficient Approach:
Calculate the maximum index that can be reached within the given steps. If the 0th Index can be reached from the maximum index by avoiding the bad index, print the result. Otherwise, repeat the procedure by decrementing the maximum index by 1.
Below are the steps:
- Calculate the maximum index that can be reached in N steps by calculating the sum of the first N natural numbers.
- Assign the value of the calculated maximum index to the Current Index.
- Keep decrementing Current Index by Step Number and Step Number by 1 until one of them becomes negative.
- After every decrement, check if the Current Index is equal to B or not. If found to be true, revert the changes made on the Current Index.
- If the Current Index reaches 0 successfully, print the current value of the maximum index as the answer.
- Otherwise, decrement the value of the maximum index by 1 and repeat from step 2.
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
void maximumIndex(int N, int B)
{
int max_index = 0;
// Calculate maximum possible
// index that can be reached
for (int i = 1; i <= N; i++) {
max_index += i;
}
int current_index = max_index, step = N;
while (1) {
// Check if current index and step
// both are greater than 0 or not
while (current_index > 0 && N > 0) {
// Decrement current_index by step
current_index -= N;
// Check if current index is
// equal to B or not
if (current_index == B) {
// Restore to previous index
current_index += N;
}
// Decrement step by one
N--;
}
// If it reaches the 0th index
if (current_index <= 0) {
// Print result
cout << max_index << endl;
break;
}
// If max index fails to
// reach the 0th index
else {
N = step;
// Store max_index - 1 in current index
current_index = max_index - 1;
// Decrement max index
max_index--;
// If current index is equal to B
if (current_index == B) {
current_index = max_index - 1;
// Decrement current index
max_index--;
}
}
}
}
// Driver Code
int main()
{
int N = 3, B = 2;
maximumIndex(N, B);
return 0;
}
// Java program for
// the above approach
import java.util.*;
class GFG{
// Function to find the maximum
// index the pointer can reach
static void maximumIndex(int N,
int B)
{
int max_index = 0;
// Calculate maximum possible
// index that can be reached
for (int i = 1; i <= N; i++)
{
max_index += i;
}
int current_index = max_index,
step = N;
while (true)
{
// Check if current index
// and step both are greater
// than 0 or not
while (current_index > 0 &&
N > 0)
{
// Decrement current_index
// by step
current_index -= N;
// Check if current index
// is equal to B or not
if (current_index == B)
{
// Restore to previous
// index
current_index += N;
}
// Decrement step by one
N--;
}
// If it reaches the 0th index
if (current_index <= 0)
{
// Print result
System.out.print(max_index + "\n");
break;
}
// If max index fails to
// reach the 0th index
else
{
N = step;
// Store max_index - 1 in
// current index
current_index = max_index - 1;
// Decrement max index
max_index--;
// If current index is
// equal to B
if (current_index == B)
{
current_index = max_index - 1;
// Decrement current index
max_index--;
}
}
}
}
// Driver Code
public static void main(String[] args)
{
int N = 3, B = 2;
maximumIndex(N, B);
}
}
// This code is contributed by gauravrajput1
# Python3 program for the above approach
# Function to find the maximum
# index the pointer can reach
def maximumIndex(N, B):
max_index = 0
# Calculate maximum possible
# index that can be reached
for i in range(1, N + 1):
max_index += i
current_index = max_index
step = N
while (1):
# Check if current index and step
# both are greater than 0 or not
while (current_index > 0 and N > 0):
# Decrement current_index by step
current_index -= N
# Check if current index is
# equal to B or not
if (current_index == B):
# Restore to previous index
current_index += N
# Decrement step by one
N -= 1
# If it reaches the 0th index
if (current_index <= 0):
# Print result
print(max_index)
break
# If max index fails to
# reach the 0th index
else:
N = step
# Store max_index - 1 in current index
current_index = max_index - 1
# Decrement max index
max_index -= 1
# If current index is equal to B
if (current_index == B):
current_index = max_index - 1
# Decrement current index
max_index -= 1
# Driver Code
if __name__ == '__main__':
N = 3
B = 2
maximumIndex(N, B)
# This code is contributed by mohit kumar 29
// 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)
{
int max_index = 0;
// Calculate maximum possible
// index that can be reached
for(int i = 1; i <= N; i++)
{
max_index += i;
}
int current_index = max_index,
step = N;
while (true)
{
// Check if current index
// and step both are greater
// than 0 or not
while (current_index > 0 &&
N > 0)
{
// Decrement current_index
// by step
current_index -= N;
// Check if current index
// is equal to B or not
if (current_index == B)
{
// Restore to previous
// index
current_index += N;
}
// Decrement step by one
N--;
}
// If it reaches the 0th index
if (current_index <= 0)
{
// Print result
Console.Write(max_index + " ");
break;
}
// If max index fails to
// reach the 0th index
else
{
N = step;
// Store max_index - 1 in
// current index
current_index = max_index - 1;
// Decrement max index
max_index--;
// If current index is
// equal to B
if (current_index == B)
{
current_index = max_index - 1;
// Decrement current index
max_index--;
}
}
}
}
// Driver code
public static void Main (String[] args)
{
int N = 3, B = 2;
maximumIndex(N, B);
}
}
// This code is contributed by offbeat
<script>
// Javascript program for the above approach
// Function to find the maximum
// index the pointer can reach
function maximumIndex( N, B)
{
var max_index = 0;
// Calculate maximum possible
// index that can be reached
for (var i = 1; i <= N; i++) {
max_index += i;
}
var current_index = max_index, step = N;
while (1) {
// Check if current index and step
// both are greater than 0 or not
while (current_index > 0 && N > 0) {
// Decrement current_index by step
current_index -= N;
// Check if current index is
// equal to B or not
if (current_index == B) {
// Restore to previous index
current_index += N;
}
// Decrement step by one
N--;
}
// If it reaches the 0th index
if (current_index <= 0) {
// Print result
document.write(max_index + "<br>");;
break;
}
// If max index fails to
// reach the 0th index
else {
N = step;
// Store max_index - 1 in current index
current_index = max_index - 1;
// Decrement max index
max_index--;
// If current index is equal to B
if (current_index == B) {
current_index = max_index - 1;
// Decrement current index
max_index--;
}
}
}
}
// Driver Code
var N = 3, B = 2;
maximumIndex(N, B);
// This code is contributed by rrrtnx.
</script>
Time Complexity: O(N2)
Auxiliary Space: O(1)
Another Efficient Approach - Keep calculating the sum (1 + 2 + 3 + .... ). Since the elements contributing to the sum is monotonically increasing so if you get the sum equals bad index then it means answer will be (sum of natural numbers - 1) else you return the sum
C++
#include <iostream>
// Function to find the maximum index 'N' such that the sum
// of natural numbers from 1 to N is equal to 'B'.
int MaximumIndex(int N, int B) {
int s = 0; // Initialize a variable 's' to keep track of the current sum.
for (int i = 1; i <= N; i++) { // Loop through natural numbers from 1 to N.
s += i; // Add the current number 'i' to the sum 's'.
if (s == B) { // If the sum 's' equals 'B', we found the answer.
// Calculate the sum of natural numbers from 1 to N using the formula.
int sum_of_natural_nos = N * (N + 1) / 2;
return sum_of_natural_nos - 1; // Return N-1 as the maximum index.
}
}
return s; // If we didn't find a match, return the current sum 's'.
}
int main() {
int N = 3;
int B = 1;
std::cout << MaximumIndex(N, B) << std::endl; // Call the function and print the result.
return 0;
}
public class Main {
// Function to find the maximum index 'N' such that the sum
// of natural numbers from 1 to N is equal to 'B'.
public static int MaximumIndex(int N, int B) {
int s = 0; // Initialize a variable 's' to keep track of the current sum.
for (int i = 1; i <= N; i++) { // Loop through natural numbers from 1 to N.
s += i; // Add the current number 'i' to the sum 's'.
if (s == B) { // If the sum 's' equals 'B', we found the answer.
// Calculate the sum of natural numbers from 1 to N using the formula.
int sum_of_natural_nos = N * (N + 1) / 2;
return sum_of_natural_nos - 1; // Return N-1 as the maximum index.
}
}
return s; // If we didn't find a match, return the current sum 's'.
}
public static void main(String[] args) {
int N = 3;
int B = 1;
System.out.println(MaximumIndex(N, B)); // Call the function and print the result.
}
}
def MaximumIndex(N, B):
s = 0
for i in range(1, N+1):
s += i
if s == B:
sum_of_natural_nos = N*(N+1)//2
return sum_of_natural_nos - 1
return s
# N = 4
# B = 6
# N = 3
# B = 2
N = 3
B = 1
print(MaximumIndex(N, B))
# This code is contributed by Swagato Chakraborty(swagatochakraborty123)
using System;
class Program
{
// Function to find the maximum index 'N' such that the sum
// of natural numbers from 1 to N is equal to 'B'.
static int MaximumIndex(int N, int B)
{
int s = 0; // Initialize a variable 's' to keep track of the current sum.
for (int i = 1; i <= N; i++) // Loop through natural numbers from 1 to N.
{
s += i; // Add the current number 'i' to the sum 's'.
if (s == B) // If the sum 's' equals 'B', we found the answer.
{
// Calculate the sum of natural numbers from 1 to N using the formula.
int sum_of_natural_nos = (N * (N + 1)) / 2;
return sum_of_natural_nos - 1; // Return N-1 as the maximum index.
}
}
return s; // If we didn't find a match, return the current sum 's'.
}
static void Main()
{
int N = 3;
int B = 1;
Console.WriteLine(MaximumIndex(N, B)); // Call the function and print the result.
}
}
// Function to find the maximum index 'N' such that the sum
// of natural numbers from 1 to N is equal to 'B'.
function maximumIndex(N, B) {
let s = 0; // Initialize a variable 's' to keep track of the current sum.
for (let i = 1; i <= N; i++) { // Loop through natural numbers from 1 to N.
s += i; // Add the current number 'i' to the sum 's'.
if (s === B) { // If the sum 's' equals 'B', we found the answer.
// Calculate the sum of natural numbers from 1 to N using the formula.
const sumOfNaturalNos = (N * (N + 1)) / 2;
return sumOfNaturalNos - 1; // Return N-1 as the maximum index.
}
}
return s; // If we didn't find a match, return the current sum 's'.
}
const N = 3;
const B = 1;
console.log(maximumIndex(N, B)); // Call the function and print the result.
Time Complexity - O(n)
Space Complexity - O(1)
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