# Maximum index a pointer can reach in N steps by avoiding a given index B – Set 3 (Binary Search)

• Difficulty Level : Easy
• Last Updated : 07 Oct, 2021

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

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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.

Efficient Approach: The naive approach discussed in this article can also be optimized by using Binary Search. If the value of maximumIndex – B exists in the sum of any last K numbers from the first N natural numbers then the maximumIndex can’t be reduced to less than equal to 0 without the jump on index B. therefore, decrement the maximum Index by and perform the above steps again. Follow the steps below to solve the problem:

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach``#include ``using` `namespace` `std;` `// Function to find the maximum index``// reachable``int` `maxIndex(``int` `N, ``int` `B)``{` `    ``// Store the answer``    ``int` `maximumIndexReached = 0;``    ``vector<``int``> Ans;` `    ``// Store the maximum index possible``    ``// i.e, N*(N-1)``    ``for` `(``int` `i = 1; i <= N; i++) {``        ``maximumIndexReached += i;``        ``Ans.push_back(i);``    ``}` `    ``reverse(Ans.begin(), Ans.end());` `    ``// Add bthe previous elements``    ``for` `(``int` `i = 1; i < (``int``)Ans.size(); i++) {``        ``Ans[i] += Ans[i - 1];``    ``}` `    ``// Run a loop``    ``while` `(maximumIndexReached) {` `        ``// Binary Search``        ``auto` `it``            ``= lower_bound(Ans.begin(), Ans.end(),``                          ``maximumIndexReached - B);``        ``if` `(it == Ans.end()) {``            ``break``;``        ``}``        ``if` `(*it == maximumIndexReached - B) {``            ``maximumIndexReached--;``        ``}``        ``else` `{``            ``break``;``        ``}``    ``}` `    ``return` `maximumIndexReached;``}` `// Driver Code``int` `main()``{``    ``int` `N = 3, B = 2;``    ``cout << maxIndex(N, B);` `    ``return` `0;``}`

## Python3

 `# Python 3 program for the above approach``from` `bisect ``import` `bisect_left` `# Function to find the maximum index``# reachable``def` `maxIndex(N, B):` `    ``# Store the answer``    ``maximumIndexReached ``=` `0``    ``Ans ``=` `[]` `    ``# Store the maximum index possible``    ``# i.e, N*(N-1)``    ``for` `i ``in` `range``(``1``, N ``+` `1``):``        ``maximumIndexReached ``+``=` `i``        ``Ans.append(i)` `    ``Ans.reverse()` `    ``# Add bthe previous elements``    ``for` `i ``in` `range``(``len``(Ans)):``        ``Ans[i] ``+``=` `Ans[i ``-` `1``]` `    ``# Run a loop``    ``while` `(maximumIndexReached):` `        ``# Binary Search``        ``it ``=` `bisect_left(Ans,``                         ``maximumIndexReached ``-` `B)``        ``if` `(it ``not` `in` `Ans):``            ``break` `        ``if` `(it ``=``=` `maximumIndexReached ``-` `B):``            ``maximumIndexReached ``-``=` `1` `        ``else``:``            ``break` `    ``return` `maximumIndexReached` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:` `    ``N ``=` `3``    ``B ``=` `2``    ``print``(maxIndex(N, B))` `    ``# This code is contributed by ukasp.`

## Javascript

 ``

Output:
`6`

Time Complexity: O(N*log N)
Auxiliary Space: O(N)

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