Half Monotonous String - Google Top Interview Questions
Problem Statement :
You are given a lowercase alphabet string s of even length consisting of lowercase alphabet letters. Return the minimum number of characters that need to be updated such that one of the following three conditions is true for all 0 ≤ i < n/2 and n/2 ≤ j < n: s[i] > s[j] or s[i] < s[j] or s[i] == s[j] Constraints 0 ≤ n ≤ 100,000 where n is the length of s Example 1 Input s = "aa" Output 0 Explanation This string already meets the s[i] == s[j] condition Example 2 Input s = "aaabba" Output 1 Explanation If we change the last "a" to "b", then we can satisfy the s[i] < s[j] condition. Example 3 Input s = "bccbba" Output 1 Explanation If we change the first "b" to "c", then we can satisfy the s[i] > s[j] condition.
Solution :
Solution in C++ :
int solve(string s) {
int len = s.length(), maxCnt = 0;
vector<int> cnt(26, 0);
for (int i = 0; i < len; i++) {
cnt[s[i] - 'a']++;
maxCnt = max(maxCnt, cnt[s[i] - 'a']);
}
int res = len - maxCnt;
for (char ch = 'b'; ch < 'z'; ch++) {
int cnt1 = 0, cnt2 = 0, cnt3 = 0, cnt4 = 0;
for (int i = 0; i < len / 2; i++) {
if (s[i] >= ch) {
cnt1++;
} else if (s[i] <= ch - 1) {
cnt2++;
}
}
for (int i = len / 2; i < len; i++) {
if (s[i] >= ch) {
cnt3++;
}
if (s[i] <= ch - 1) {
cnt4++;
}
}
res = min({res, cnt1 + cnt4, cnt2 + cnt3});
}
return res;
}
Solution in Java :
import java.util.*;
class Solution {
public int solve(String s) {
int N = s.length();
int[] A = new int[26];
int[] B = new int[26];
for (int i = 0; i < N; i++) {
int c = s.charAt(i) - 'a';
if (i < N / 2)
A[c] += 1;
else
B[c] += 1;
}
int ans = Integer.MAX_VALUE;
// fix all characters in A to be greater than i and all the characters in B to be less than
// or equal to i.
for (int i = 0; i < 26; i++) {
int fix = 0;
for (int j = 0; j <= i; j++) fix += A[j];
for (int j = i + 1; j < 26; j++) fix += B[j];
ans = Math.min(ans, fix);
}
// fix all characters in A to be less than i and all the characters in B to be greater than
// or equal to i.
for (int i = 0; i < 26; i++) {
int fix = 0;
for (int j = i; j < 26; j++) fix += A[j];
for (int j = 0; j < i; j++) fix += B[j];
ans = Math.min(ans, fix);
}
// fix all characters in A and B to equal i.
for (int i = 0; i < 26; i++) {
int fix = N - (A[i] + B[i]);
ans = Math.min(ans, fix);
}
return ans;
}
}
Solution in Python :
class Solution:
def solve(self, s):
n = len(s)
left = Counter(s[: n >> 1])
right = Counter(s[n >> 1 :])
ans = n
for pivot in ascii_lowercase:
# All same
ans = min(ans, n - left[pivot] - right[pivot])
# left <= pivot < right
good = sum(left[c] for c in left if c <= pivot)
good += sum(right[c] for c in right if c > pivot)
ans = min(ans, n - good)
# right <= pivot < left
good = sum(left[c] for c in left if c > pivot)
good += sum(right[c] for c in right if c <= pivot)
ans = min(ans, n - good)
return ans
View More Similar Problems
Merge two sorted linked lists
This challenge is part of a tutorial track by MyCodeSchool Given pointers to the heads of two sorted linked lists, merge them into a single, sorted linked list. Either head pointer may be null meaning that the corresponding list is empty. Example headA refers to 1 -> 3 -> 7 -> NULL headB refers to 1 -> 2 -> NULL The new list is 1 -> 1 -> 2 -> 3 -> 7 -> NULL. Function Description C
View Solution →Get Node Value
This challenge is part of a tutorial track by MyCodeSchool Given a pointer to the head of a linked list and a specific position, determine the data value at that position. Count backwards from the tail node. The tail is at postion 0, its parent is at 1 and so on. Example head refers to 3 -> 2 -> 1 -> 0 -> NULL positionFromTail = 2 Each of the data values matches its distance from the t
View Solution →Delete duplicate-value nodes from a sorted linked list
This challenge is part of a tutorial track by MyCodeSchool You are given the pointer to the head node of a sorted linked list, where the data in the nodes is in ascending order. Delete nodes and return a sorted list with each distinct value in the original list. The given head pointer may be null indicating that the list is empty. Example head refers to the first node in the list 1 -> 2 -
View Solution →Cycle Detection
A linked list is said to contain a cycle if any node is visited more than once while traversing the list. Given a pointer to the head of a linked list, determine if it contains a cycle. If it does, return 1. Otherwise, return 0. Example head refers 1 -> 2 -> 3 -> NUL The numbers shown are the node numbers, not their data values. There is no cycle in this list so return 0. head refer
View Solution →Find Merge Point of Two Lists
This challenge is part of a tutorial track by MyCodeSchool Given pointers to the head nodes of 2 linked lists that merge together at some point, find the node where the two lists merge. The merge point is where both lists point to the same node, i.e. they reference the same memory location. It is guaranteed that the two head nodes will be different, and neither will be NULL. If the lists share
View Solution →Inserting a Node Into a Sorted Doubly Linked List
Given a reference to the head of a doubly-linked list and an integer ,data , create a new DoublyLinkedListNode object having data value data and insert it at the proper location to maintain the sort. Example head refers to the list 1 <-> 2 <-> 4 - > NULL. data = 3 Return a reference to the new list: 1 <-> 2 <-> 4 - > NULL , Function Description Complete the sortedInsert function
View Solution →