Given a string str, the task is to find the minimum numbers of substrings that the given string S can be split into, such that each substring is monotonously increasing or decreasing.
Examples:
Input: str = “abcdcba”
Output: 2
Explanation:
The string can be split into a minimum of 2 monotonous substrings {“abcd”(increasing), “cba”(decreasing)}
Input: str = “aeccdhba”
Output: 3
Explanation:
The generated substrings are {“ae”, “ccdh”, “ba”}
Approach: Follow the steps below to solve the problem:
- Initialize a variable ongoing = ‘N’ to keep track of order of current sequence.
- Iterate over the string and for each character, follow the steps below:
- If ongoing == ‘N’:
- If curr_character < prev_character then update ongoing with D(Non-Increasing).
- Otherwise, if curr_character > prev_character, then update ongoing with I(Non-Decreasing).
- Otherwise, update ongoing with N(neither Non-Increasing nor Non-Decreasing).
- Otherwise, if ongoing == ‘I’:
- If curr_character > prev_character then update ongoing with I.
- Otherwise, if curr_character < prev_character then update ongoing with N and increment answer.
- Otherwise, update ongoing with I.
- else do the following steps:
- If curr_character < prev_character then update ongoing with D.
- Otherwise, if curr_character > prev_character then update ongoing with N and increment answer.
- Otherwise, update ongoing with D.
- Finally, print answer+1 is the required answer.
Below is the implementation of the above approach:
C++
// C++ program to implement// the above approach#include<bits/stdc++.h>using namespace std;// Function to return final resultint minReqSubstring(string s, int n){ // Initialize variable to keep // track of ongoing sequence char ongoing = 'N'; int count = 0, l = s.size(); for(int i = 1; i < l; i++) { // If current sequence is neither // increasing nor decreasing if (ongoing == 'N') { // If prev char is greater if (s[i] < s[i - 1]) { ongoing = 'D'; } // If prev char is same else if (s[i] == s[i - 1]) { ongoing = 'N'; } // Otherwise else { ongoing = 'I'; } } // If current sequence // is Non-Decreasing else if (ongoing == 'I') { // If prev char is smaller if (s[i] > s[i - 1]) { ongoing = 'I'; } // If prev char is same else if (s[i] == s[i - 1]) { // Update ongoing ongoing = 'I'; } // Otherwise else { // Update ongoing ongoing = 'N'; // Increase count count += 1; } } // If current sequence // is Non-Increasing else { // If prev char is greater, // then update ongoing with D if (s[i] < s[i - 1]) { ongoing = 'D'; } // If prev char is equal, then // update current with D else if (s[i] == s[i - 1]) { ongoing = 'D'; } // Otherwise, update ongoing // with N and increment count else { ongoing = 'N'; count += 1; } } } // Return count+1 return count + 1;}// Driver Codeint main(){ string S = "aeccdhba"; int n = S.size(); cout << (minReqSubstring(S, n)); return 0;}// This code is contributed by Amit Katiyar |
Java
// Java Program to implement// the above approachimport java.util.*;class GFG { // Function to return final result static int minReqSubstring(String s, int n) { // Initialize variable to keep // track of ongoing sequence char ongoing = 'N'; int count = 0, l = s.length(); for (int i = 1; i < l; i++) { // If current sequence is neither // increasing nor decreasing if (ongoing == 'N') { // If prev char is greater if (s.charAt(i) < s.charAt(i - 1)) { ongoing = 'D'; } // If prev char is same else if (s.charAt(i) == s.charAt(i - 1)) { ongoing = 'N'; } // Otherwise else { ongoing = 'I'; } } // If current sequence // is Non-Decreasing else if (ongoing == 'I') { // If prev char is smaller if (s.charAt(i) > s.charAt(i - 1)) { ongoing = 'I'; } // If prev char is same else if (s.charAt(i) == s.charAt(i - 1)) { // Update ongoing ongoing = 'I'; } // Otherwise else { // Update ongoing ongoing = 'N'; // Increase count count += 1; } } // If current sequence // is Non-Increasing else { // If prev char is greater, // then update ongoing with D if (s.charAt(i) < s.charAt(i - 1)) { ongoing = 'D'; } // If prev char is equal, then // update current with D else if (s.charAt(i) == s.charAt(i - 1)) { ongoing = 'D'; } // Otherwise, update ongoing // with N and increment count else { ongoing = 'N'; count += 1; } } } // Return count+1 return count + 1; } // Driver Code public static void main(String[] args) { String S = "aeccdhba"; int n = S.length(); System.out.print( minReqSubstring(S, n)); }} |
Python3
# Python3 program to implement # the above approach # Function to return final result def minReqSubstring(s, n): # Initialize variable to keep # track of ongoing sequence ongoing = 'N' count, l = 0, len(s) for i in range(1, l): # If current sequence is neither # increasing nor decreasing if ongoing == 'N': # If prev char is greater if s[i] < s[i - 1]: ongoing = 'D' # If prev char is same elif s[i] == s[i - 1]: ongoing = 'N' # Otherwise else: ongoing = 'I' # If current sequence # is Non-Decreasing elif ongoing == 'I': # If prev char is smaller if s[i] > s[i - 1]: ongoing = 'I' # If prev char is same elif s[i] == s[i - 1]: # Update ongoing ongoing = 'I' # Otherwise else: # Update ongoing ongoing = 'N' # Increase count count += 1 # If current sequence # is Non-Increasing else: # If prev char is greater, # then update ongoing with D if s[i] < s[i - 1]: ongoing = 'D' # If prev char is equal, then # update current with D elif s[i] == s[i - 1]: ongoing = 'D' # Otherwise, update ongoing # with N and increment count else: ongoing = 'N' count += 1 # Return count + 1 return count + 1# Driver codeS = "aeccdhba"n = len(S)print(minReqSubstring(S, n))# This code is contributed by Stuti Pathak |
C#
// C# Program to implement// the above approachusing System;class GFG{ // Function to return readonly result static int minReqSubstring(String s, int n) { // Initialize variable to keep // track of ongoing sequence char ongoing = 'N'; int count = 0, l = s.Length; for (int i = 1; i < l; i++) { // If current sequence is neither // increasing nor decreasing if (ongoing == 'N') { // If prev char is greater if (s[i] < s[i - 1]) { ongoing = 'D'; } // If prev char is same else if (s[i] == s[i - 1]) { ongoing = 'N'; } // Otherwise else { ongoing = 'I'; } } // If current sequence // is Non-Decreasing else if (ongoing == 'I') { // If prev char is smaller if (s[i] > s[i - 1]) { ongoing = 'I'; } // If prev char is same else if (s[i] == s[i - 1]) { // Update ongoing ongoing = 'I'; } // Otherwise else { // Update ongoing ongoing = 'N'; // Increase count count += 1; } } // If current sequence // is Non-Increasing else { // If prev char is greater, // then update ongoing with D if (s[i] < s[i - 1]) { ongoing = 'D'; } // If prev char is equal, then // update current with D else if (s[i] == s[i - 1]) { ongoing = 'D'; } // Otherwise, update ongoing // with N and increment count else { ongoing = 'N'; count += 1; } } } // Return count+1 return count + 1; } // Driver Code public static void Main(String[] args) { String S = "aeccdhba"; int n = S.Length; Console.Write(minReqSubstring(S, n)); }}// This code is contributed by Rohit_ranjan |
Javascript
<script>// javascript program for the// above approach // Function to return final result function minReqSubstring(s, n) { // Initialize variable to keep // track of ongoing sequence let ongoing = 'N'; let count = 0, l = s.length; for (let i = 1; i < l; i++) { // If current sequence is neither // increasing nor decreasing if (ongoing == 'N') { // If prev char is greater if (s[i] < s[i - 1]) { ongoing = 'D'; } // If prev char is same else if (s[i] == s[i - 1]) { ongoing = 'N'; } // Otherwise else { ongoing = 'I'; } } // If current sequence // is Non-Decreasing else if (ongoing == 'I') { // If prev char is smaller if (s[i] > s[i - 1]) { ongoing = 'I'; } // If prev char is same else if (s[i] == s[i - 1]) { // Update ongoing ongoing = 'I'; } // Otherwise else { // Update ongoing ongoing = 'N'; // Increase count count += 1; } } // If current sequence // is Non-Increasing else { // If prev char is greater, // then update ongoing with D if (s[i] < s[i - 1]) { ongoing = 'D'; } // If prev char is equal, then // update current with D else if (s[i] == s[i - 1]) { ongoing = 'D'; } // Otherwise, update ongoing // with N and increment count else { ongoing = 'N'; count += 1; } } } // Return count+1 return count + 1; } // Driver Code let S = "aeccdhba"; let n = S.length; document.write( minReqSubstring(S, n));// This code is contributed by target_2.</script> |
Output:
3
Time Complexity: O(N)
Auxiliary Space: O(1)
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