Monotonous String Groups πŸŽƒ - Google Top Interview Questions

Problem Statement :

You are given a lowercase alphabet string s. Return the minimum numbers of contiguous substrings in which s must be broken into such that each substring is either non-increasing or non-decreasing.

For example, "acccf" is a non-decreasing string, and "bbba" is a non-increasing string.


n ≀ 100,000 where n is the length of s

Example 1


s = "abcdcba"




We can break s into "abcd" + "cba"

Example 2


s = "zzz"




We can break s into just "zzz"

Solution :


                        Solution in C++ :

int solve(string s) {
    int ret = 0;
    for (int i = 0; i < s.size();) {
        int j = i + 1;
        bool noninc = true;
        bool nondec = true;
        while (j < s.size()) {
            if (s[j] > s[j - 1]) noninc = false;
            if (s[j] < s[j - 1]) nondec = false;
            if (!noninc && !nondec) break;
        i = j;
    return ret;

                        Solution in Java :

import java.util.*;

class Solution {
    public int solve(String s) {
        // If empty string, there are no substrings
        if (s.length() < 1) {
            return 0;

        // Counter to keep track of number of substrings
        int count = 1;

        // Number to indicate whether current substring is
        // non-decreasing, non-increasing, or neither decreasing nor
        // increasing (staying the same).
        int current = 0;

        for (int i = 1; i < s.length(); i++) {
            // Find out out whether behavior is decreasing, staying the same
            // or increasing between current character and previous character
            int diff = s.charAt(i) - s.charAt(i - 1);

            // If the current substring's behavior has not been determined yet
            // (can still either be non-decreasing or non-increasing)
            if (current == 0) {
                // Set the current string's behavior to non-increasing
                if (diff < 0) {
                    current = -1;
                    // Set the current string's behavior to non-decreasing
                } else if (diff > 0) {
                    current = 1;

                // Note that if diff is 0, the current string's behavior can be
                // determined at a later time

                // If the current substring's behavior has already been determined
                // (either non-decreasing or non-increasing)
            } else {
                // If the behavior between this index and the previous index does
                // not match the behavior of the current substring, it is time
                // to start a new substring
                if (diff < 0 && current == 1 || diff > 0 && current == -1) {
                    current = 0;

        return count;

                        Solution in Python : 
class Solution:
    def solve(self, s):
        if not s:
            return 0
        count = 1
        increasing = None
        for i in range(len(s) - 1):
            if increasing is None and s[i] != s[i + 1]:
                increasing = s[i] < s[i + 1]
            elif (increasing and s[i] > s[i + 1]) or (increasing is False and s[i] < s[i + 1]):
                count += 1
                increasing = None
        return count

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