Problem:
Delete Columns to Make SortedWe are given an array A of N lowercase letter strings, all of the same length.
Now, we may choose any set of deletion indices, and for each string, we delete all the characters in those indices.
For example, if we have an array A = ["abcdef","uvwxyz"] and deletion indices {0, 2, 3}, then the final array after deletions is ["bef", "vyz"], and the remaining columns of A are ["b","v"], ["e","y"], and ["f","z"]. (Formally, the c-th column is [A[0][c], A[1][c], ..., A[A.length-1][c]]).
Suppose we chose a set of deletion indices D such that after deletions, each remaining column in A is in non-decreasing sorted order.
Return the minimum possible value of D.length.
Example 1:
Input: A = ["cba","daf","ghi"]
Output: 1
Explanation:
After choosing D = {1}, each column ["c","d","g"] and ["a","f","i"] are in non-decreasing sorted order.
If we chose D = {}, then a column ["b","a","h"] would not be in non-decreasing sorted order.
Example 2:
Input: A = ["a","b"]
Output: 0
Explanation: D = {}
Example 3:
Input: A = ["zyx","wvu","tsr"]
Output: 3
Explanation: D = {0, 1, 2}
Constraints:
1 <= A.length <= 100
1 <= A[i].length <= 1000
Solution:
This problem has a long description, but boiling down to remove the columns with decreasing order.
class Solution {
public static void main(String...args) {
String[] arr = new String[]{"cba","daf","ghi"};
int result = solve(arr);
System.out.println(result);
}
private static int solve(String[] arr) {
int counter = 0;
for(int i = 0; i < arr[0].length(); i++) {
for(int j = 0; j < arr.length - 1; j++) {
if(arr[j].charAt(i) > arr[j+1].charAt(i)) {
counter++;
break;
}
}
}
return counter;
}
}
public static void main(String...args) {
String[] arr = new String[]{"cba","daf","ghi"};
int result = solve(arr);
System.out.println(result);
}
private static int solve(String[] arr) {
int counter = 0;
for(int i = 0; i < arr[0].length(); i++) {
for(int j = 0; j < arr.length - 1; j++) {
if(arr[j].charAt(i) > arr[j+1].charAt(i)) {
counter++;
break;
}
}
}
return counter;
}
}
The time complexity is O(A) which is the length of the string multiply the total number of strings, the space cost is O(1).