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.

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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).

C++

 // CPP program to find the maximum segment // value after putting k breaks. #include 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

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.

Output:

Maximum number = 87

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