Maximum segment value after putting k breakpoints in a number

Given a large number as string s and an integer k which denotes the number of breakpoints we must put in the number k <= string length. The task is to find maximum segment value after putting exactly k breakpoints. 

Examples:

Input : s = "8754", k = 2
Output : Maximum number = 87
Explanation : We need to two breakpoints. After
putting the breakpoints, we get following options
8 75 4
87 5 4
The maximum segment value is 87.

Input : s = "999", k = 1 
Output : Maximum Segment Value = 99
Explanation : We need to one breakpoint. After
putting the breakpoint, we either get 99,9 or
9,99.

One important observation is, the maximum would always be of length “string-length – k” which is the maximum value of any segment. Considering the fact, problem becomes like sliding window problem means we need to find maximum of all substrings of size (string-length – k). 

Implementation:

C++

// CPP program to find the maximum segment
// value after putting k breaks.
#include <bits/stdc++.h>
using namespace std;

// Function to Find Maximum Number
int findMaxSegment(string &s, int k) {

  // Maximum segment length
  int seg_len = s.length() - k;

  // Find value of first segment of seg_len
  int res = 0;
  for (int i=0; i<seg_len; i++)
     res = res * 10 + (s[i] - '0');

  // Find value of remaining segments using sliding
  // window
  int seg_len_pow = pow(10, seg_len-1);
  int curr_val = res;
  for (int i = 1; i <= (s.length() - seg_len); i++) {

    // To find value of current segment, first remove
    // leading digit from previous value
    curr_val = curr_val - (s[i-1]- '0')*seg_len_pow;

    // Then add trailing digit
    curr_val = curr_val*10 + (s[i+seg_len-1]- '0');

    res = max(res, curr_val);
  }
  return res;
}

// Driver's Function
int main() {
  string s = "8754";
  int k = 2;
  cout << "Maximum number = " << findMaxSegment(s, k);
  return 0;
}

Java

// Java program to find the maximum segment
// value after putting k breaks.
class GFG {

    // Function to Find Maximum Number
    static int findMaxSegment(String s, int k) {

        // Maximum segment length
        int seg_len = s.length() - k;

        // Find value of first segment of seg_len
        int res = 0;

        for (int i = 0; i < seg_len; i++)
            res = res * 10 + (s.charAt(i) - '0');

        // Find value of remaining segments using
        // sliding window
        int seg_len_pow = (int)Math.pow(10,
                                    seg_len - 1);
        int curr_val = res;

        for (int i = 1;
             i <= (s.length() - seg_len); i++) {

            // To find value of current segment,
            // first remove leading digit from
            // previous value
            curr_val = curr_val -
            (s.charAt(i - 1) - '0') * seg_len_pow;

            // Then add trailing digit
            curr_val = curr_val * 10 +
               (s.charAt(i + seg_len - 1) - '0');

            res = Math.max(res, curr_val);
        }

        return res;
    }

    // Driver code
    public static void main(String[] args) {

        String s = "8754";
        int k = 2;

        System.out.print("Maximum number = "
                        + findMaxSegment(s, k));
    }
}

// This code is contributed by Anant Agarwal.

Python3

# Python3 program to find the maximum segment
# value after putting k breaks.

# Function to Find Maximum Number
def findMaxSegment(s, k):

    # Maximum segment length
    seg_len = len(s) - k

    # Find value of first segment of seg_len
    res = 0
    for i in range(seg_len):
        res = res * 10 + (ord(s[i]) - ord('0'))

    # Find value of remaining segments
    # using sliding window
    seg_len_pow = pow(10, seg_len - 1)
    curr_val = res
    for i in range(1, len(s) - seg_len):

        # To find value of current segment,
        # first remove leading digit from
        # previous value
        curr_val = curr_val - (ord(s[i - 1])-
                               ord('0')) * seg_len_pow

        # Then add trailing digit
        curr_val = (curr_val * 10 +
                   (ord(s[i + seg_len - 1]) - ord('0')))

        res = max(res, curr_val)
    return res

# Driver Code
if __name__ == '__main__':
    s = "8754"
    k = 2
    print("Maximum number = ",
         findMaxSegment(s, k))

# This code is contributed by PranchalK

C#

// C# program to find the maximum segment
// value after putting k breaks.
using System;

class GFG {

    // Function to Find Maximum Number
    static int findMaxSegment(string s, int k) {

        // Maximum segment length
        int seg_len = s.Length - k;

        // Find value of first segment of seg_len
        int res = 0;

        for (int i = 0; i < seg_len; i++)
            res = res * 10 + (s[i] - '0');

        // Find value of remaining segments using
        // sliding window
        int seg_len_pow = (int)Math.Pow(10,
                                    seg_len - 1);
        int curr_val = res;

        for (int i = 1;
            i <= (s.Length- seg_len); i++) {

            // To find value of current segment,
            // first remove leading digit from
            // previous value
            curr_val = curr_val -
            (s[i - 1] - '0') * seg_len_pow;

            // Then add trailing digit
            curr_val = curr_val * 10 +
            (s[i + seg_len - 1] - '0');

            res = Math.Max(res, curr_val);
        }

        return res;
    }

    // Driver code
    public static void Main() {

        String s = "8754";
        int k = 2;

        Console.WriteLine("Maximum number = "
                        + findMaxSegment(s, k));
    }
}

// This code is contributed by vt_m.

Javascript

<script>

// JavaScript program to find the maximum segment
// value after putting k breaks.

// Function to Find Maximum Number
function findMaxSegment(s, k){

    // Maximum segment length
    let seg_len = s.length - k

    // Find value of first segment of seg_len
    let res = 0
    for(let i=0;i<seg_len;i++)
        res = res * 10 + (s.charCodeAt(i) - '0'.charCodeAt(0))

    // Find value of remaining segments
    // using sliding window
    let seg_len_pow = Math.pow(10, seg_len - 1)
    let curr_val = res
    for(let i = 1;i< s.length - seg_len;i++){

        // To find value of current segment,
        // first remove leading digit from
        // previous value
        curr_val = curr_val - (s.charCodeAt(i - 1)-'0'.charCodeAt(0)) * seg_len_pow

        // Then add trailing digit
        curr_val = (curr_val * 10 +
                (s.charCodeAt(i + seg_len - 1) - '0'.charCodeAt(0)))

        res = Math.max(res, curr_val)
    }
    return res
}

// Driver Code

let s = "8754"
let k = 2
document.writea("Maximum number = ",findMaxSegment(s, k))

// This code is contributed by shinjanpatra

</script>

Output

Maximum number = 87

Time complexity : O(K) 
Auxiliary Space : O(1)