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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 <bits/stdc++.h>
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




<script>
        // JavaScript Program to implement
        // the above approach
        function lowerBound(Ans, target)
        {
            const targetRange = [-1, -1]
            const leftIdx = extremeInsertionIndex(Ans, target, true)
            if (leftIdx === Ans.length || Ans[leftIdx] != target)
                return targetRange
            targetRange[0] = leftIdx
            targetRange[1] = extremeInsertionIndex(Ans, target, false) - 1
            return targetRange
            function extremeInsertionIndex(Ans, target, left) {
                let lo = 0,
                    hi = Ans.length
                while (lo < hi) {
                    const mid = lo + Math.floor((hi - lo) / 2)
                    if (Ans[mid] > target || (left && target === Ans[mid]))
                        hi = mid
                    else
                        lo = mid + 1
                }
                return lo
            }
        }
         
        // Function to find the maximum index
        // reachable
        function maxIndex(N, B)
        {
         
            // Store the answer
            let maximumIndexReached = 0;
            let Ans = [];
             
            // Store the maximum index possible
            // i.e, N*(N-1)
            for (let i = 1; i <= N; i++) {
                maximumIndexReached += i;
                Ans.push(i);
            }
            Ans.reverse();
             
            // Add bthe previous elements
            for (let i = 1; i < Ans.length; i++) {
                Ans[i] += Ans[i - 1];
            }
             
            // Run a loop
            while (maximumIndexReached)
            {
             
                // Binary Search
                let it
                    = lowerBound(Ans, maximumIndexReached - B);
                if (it == -1) {
                    break;
                }
                if (it == maximumIndexReached - B) {
                    maximumIndexReached--;
                }
                else {
                    break;
                }
            }
            return maximumIndexReached;
        }
         
        // Driver Code
        let N = 3, B = 2;
        document.write(maxIndex(N, B));
         
// This code is contributed by Potta Lokesh
    </script>

 
 

Output: 
6

 

 

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

 




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