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Count columns to be deleted to make each row sorted

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Given an array arr[] of strings of same length, the task is to count the number of columns to be deleted so that all the rows are lexicographically sorted. 

Examples:

Input: arr[] = {“hello”, “geeks”} 
Output: 1 Deleting column 1 (index 0) Now both strings are sorted in lexicographical order i.e. “ello” and “eeks”. 

Input: arr[] = {“xyz”, “lmn”, “pqr”} 
Output: 0 All rows are already sorted lexicographically.

Approach 1: 

Convert the given Vector/array of string into a char grid of n*m as n will be the number of row (number of string in vector/array) and m will be there size of string/length.then we will formally traverse over all the row and column and check for sorted or not if they are then continue. if they are not then we will save its column number and at the end we will return number of column that need to remove in order to make all the rows lexicographically sorted. 

C++




#include <bits/stdc++.h>
#define int long long
using namespace std;
int n, m;
bool test(int i, int j, vector<vector<char> >& v)
{
    vector<char> x;
    for (int k = j; k < m; k++) {
        x.push_back(v[i][k]);
    }
    if (is_sorted(x.begin(), x.end())) {
        return 1;
    }
    return 0;
}
void solve(vector<string>& strs)
{
    //    std::cout << "/* message */" << '\n';
    n = strs.size();
    m = strs[0].size();
    vector<vector<char> > v(n, vector<char>(m));
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            for (auto it : strs[i])
                v[i][j++] = it;
        }
    }
    // for(auto it:v){
    //     for(auto vt:it)cout<<vt<<" ";
    //     cout<<endl;
    // }
    int res = 0;
    for (int i = 0; i < n; i++) {
        int cnt = res;
        if (is_sorted(v[i].begin(), v[i].end())) {
            ++cnt;
            continue;
        }
        for (int j = cnt; j < m; j++) {
            if (test(i, j, v)) {
                break;
            }
            else
                ++cnt;
        }
        res = cnt;
    }
    cout << res << endl;
}
// Driver Code
signed main()
{
    vector<string> v{ "hello", "geeks" };
    vector<string> x{ "xyz", "lmn", "pqr" };
    // cout<<solve(v)<<endl;
    solve(v);
    solve(x);
}


Java




// Java code for the above approach
import java.util.*;
 
public class Main {
  static int n, m;
 
  static boolean test(int i, int j, char[][] v) {
    List<Character> x = new ArrayList<>();
    for (int k = j; k < m; k++) {
      x.add(v[i][k]);
    }
    char[] temp = new char[x.size()];
    for (int k = 0; k < x.size(); k++) {
      temp[k] = x.get(k);
    }
    if (isSorted(temp)) {
      return true;
    }
    return false;
  }
 
  static boolean isSorted(char[] arr) {
    for (int i = 0; i < arr.length - 1; i++) {
      if (arr[i] > arr[i + 1]) {
        return false;
      }
    }
    return true;
  }
 
  static void solve(String[] strs) {
    n = strs.length;
    m = strs[0].length();
    char[][] v = new char[n][m];
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        v[i][j] = strs[i].charAt(j);
      }
    }
    int res = 0;
    for (int i = 0; i < n; i++) {
      int cnt = res;
      if (isSorted(v[i])) {
        cnt++;
        continue;
      }
      for (int j = cnt; j < m; j++) {
        if (test(i, j, v)) {
          break;
        } else {
          cnt++;
        }
      }
      res = cnt;
    }
    System.out.println(res);
  }
 
  public static void main(String[] args) {
    String[] v = { "hello", "geeks" };
    String[] x = { "xyz", "lmn", "pqr" };
    solve(v);
    solve(x);
  }
}
 
// This code is contributed by pradeepkumarppk2003


Python3




import itertools
n, m = 0,0
def test(i, j, v):
    x = []
    for k in range(j, m):
        x.append(v[i][k])
    if(sorted(x) == x):
        return True
    return False
 
def solve(strs):
    global n, m
    n = len(strs)
    m = len(strs[0])
    v = [[0 for j in range(m)] for i in range(n)]
    for i in range(n):
        for j in range(m):
            for k, val in enumerate(strs[i]):
                v[i][j+k] = val
    res = 0
    for i in range(n):
        cnt = res
        if(sorted(v[i]) == v[i]):
            cnt += 1
            continue
        for j in range(cnt, m):
            if(test(i, j, v)):
                break
            else:
                cnt += 1
        res = cnt
    print(res)
# Driver Code
if __name__ == "__main__":
    v = ["hello", "geeks"]
    x = ["xyz", "lmn", "pqr"]
    solve(v)
    solve(x)


C#




// C# code for the above approach
using System;
using System.Collections.Generic;
 
public class MainClass {
    static int n, m;
 
    static bool Test(int i, int j, char[][] v)
    {
        List<char> x = new List<char>();
        for (int k = j; k < m; k++) {
            x.Add(v[i][k]);
        }
        char[] temp = new char[x.Count];
        for (int k = 0; k < x.Count; k++) {
            temp[k] = x[k];
        }
        if (IsSorted(temp)) {
            return true;
        }
        return false;
    }
 
    static bool IsSorted(char[] arr)
    {
        for (int i = 0; i < arr.Length - 1; i++) {
            if (arr[i] > arr[i + 1]) {
                return false;
            }
        }
        return true;
    }
 
    static void Solve(string[] strs)
    {
        n = strs.Length;
        m = strs[0].Length;
        char[][] v = new char[n][];
        for (int i = 0; i < n; i++) {
            v[i] = new char[m];
            for (int j = 0; j < m; j++) {
                v[i][j] = strs[i][j];
            }
        }
        int res = 0;
        for (int i = 0; i < n; i++) {
            int cnt = res;
            if (IsSorted(v[i])) {
                cnt++;
                continue;
            }
            for (int j = cnt; j < m; j++) {
                if (Test(i, j, v)) {
                    break;
                }
                else {
                    cnt++;
                }
            }
            res = cnt;
        }
        Console.WriteLine(res);
    }
 
    public static void Main()
    {
        string[] v = { "hello", "geeks" };
        string[] x = { "xyz", "lmn", "pqr" };
        Solve(v);
        Solve(x);
    }
}


Javascript




function test(i, j, v) {
    let x = [];
    for (let k = j; k < m; k++) {
        x.push(v[i][k]);
    }
    if (x.every((val, index, arr) => !index || val >= arr[index - 1])) {
        return true;
    }
    return false;
}
 
function solve(strs) {
    n = strs.length;
    m = strs[0].length;
    let v = new Array(n).fill().map(() => new Array(m).fill());
    for (let i = 0; i < n; i++) {
        for (let j = 0; j < m; j++) {
            v[i][j] = strs[i].charAt(j);
        }
    }
    let res = 0;
    for (let i = 0; i < n; i++) {
        let cnt = res;
        if (v[i].every((val, index, arr) => !index || val >= arr[index - 1])) {
            cnt++;
            continue;
        }
        for (let j = cnt; j < m; j++) {
            if (test(i, j, v)) {
                break;
            } else {
                cnt++;
            }
        }
        res = cnt;
    }
    console.log(res);
}
 
let v = ["hello", "geeks"];
let x = ["xyz", "lmn", "pqr"];
 
solve(v);
solve(x);


Output

1
0

Approach 2: 

The idea of the problem is to find the columns to keep, instead of columns to delete. Finally we return the difference of the counted value with the length of the string. Now, let’s say we keep the first column C1. The next column C2 we must have all rows lexicographically sorted i.e. C1[i] <= C2[i] for all valid values of i and we say that we have deleted all columns between C1 and C2.

Below is the implementation of the above approach: 

C++




#include <bits/stdc++.h>
using namespace std;
 
// Function to find minimum columns to be deleted
int deleteColumns(vector<string>& A)
{
    // Length of each string
    int l = A[0].length();
 
    // Initialize dp array
    vector<int> dp(l, 1);
 
    for (int i = l - 2; i >= 0; i--) {
        for (int j = i + 1; j < l; j++) {
            bool flag = true;
            for (auto row : A) {
                if (row[i] > row[j]) {
                    flag = false;
                    break;
                }
            }
            if (flag) {
                dp[i] = max(dp[i], 1 + dp[j]);
            }
        }
    }
 
    // Return result
    return l - *max_element(dp.begin(), dp.end());
}
 
// Driver Code
int main()
{
    vector<string> arr = { "hello", "geeks" };
    vector<string> arr2 = { "xyz", "lmn", "pqr" };
 
    // Function call to print required answer
    cout << deleteColumns(arr) << endl;
    cout << deleteColumns(arr2) << endl;
 
    return 0;
}


Java




// Java program for the above approach
import java.util.*;
class Main {
 
  // Function to find minimum columns to be deleted
  static int deleteColumns(List<String> A) {
 
    // Length of each string
    int l = A.get(0).length();
 
    // Initialize dp array
    int[] dp = new int[l];
    Arrays.fill(dp, 1);
 
    for (int i = l - 2; i >= 0; i--) {
      for (int j = i + 1; j < l; j++) {
        boolean flag = true;
        for (String row : A) {
          if (row.charAt(i) > row.charAt(j)) {
            flag = false;
            break;
          }
        }
        if (flag) {
          dp[i] = Math.max(dp[i], 1 + dp[j]);
        }
      }
    }
 
    // Return result
    return l - Arrays.stream(dp).max().getAsInt();
  }
 
  // Driver Code
  public static void main(String[] args) {
    List<String> arr = Arrays.asList("hello", "geeks");
    List<String> arr2 = Arrays.asList("xyz", "lmn", "pqr");
 
    // Function call to print required answer
    System.out.println(deleteColumns(arr));
    System.out.println(deleteColumns(arr2));
  }
}
 
// This code is contributed by Prince


Python3




# Python3 implementation of the approach
 
# Function to find minimum columns to be deleted
def deleteColumns(A):
 
    # Length of each string
    l = len(A[0])
 
    # Initialize dp array
    dp = [1] * l
 
    for i in range(l - 2, -1, -1):
        for j in range(i + 1, l):
            if all(row[i] <= row[j] for row in A):
                dp[i] = max(dp[i], 1 + dp[j])
 
    # Return result
    return l - max(dp)
 
 
# Driver Code
arr = ["hello", "geeks"]
arr2 = ["xyz", "lmn", "pqr"]
 
# Function call to print required answer
print(deleteColumns(arr)+"<br>")
print(deleteColumns(arr2))


C#




// C# program for the above approach
using System;
using System.Collections.Generic;
using System.Linq;
 
public class MainClass {
    // Function to find minimum columns to be deleted
    static int DeleteColumns(List<string> A) {
 
        // Length of each string
        int l = A[0].Length;
 
        // Initialize dp array
        int[] dp = new int[l];
        Array.Fill(dp, 1);
 
        for (int i = l - 2; i >= 0; i--) {
            for (int j = i + 1; j < l; j++) {
                bool flag = true;
                foreach (string row in A) {
                    if (row[i] > row[j]) {
                        flag = false;
                        break;
                    }
                }
                if (flag) {
                    dp[i] = Math.Max(dp[i], 1 + dp[j]);
                }
            }
        }
 
        // Return result
        return l - dp.Max();
    }
 
    public static void Main() {
        List<string> arr = new List<string>() { "hello", "geeks" };
        List<string> arr2 = new List<string>() { "xyz", "lmn", "pqr" };
 
        // Function call to print required answer
        Console.WriteLine(DeleteColumns(arr));
        Console.WriteLine(DeleteColumns(arr2));
    }
}
 
 
// This code is contributed by codebraxnzt


Javascript




//JS code for the above approach
 
// Function to find minimum columns to be deleted
function deleteColumns(A) {
    // Length of each string
    let l = A[0].length;
 
    // Initialize dp array
    let dp = Array(l).fill(1);
 
    for (let i = l - 2; i >= 0; i--) {
        for (let j = i + 1; j < l; j++) {
            if (A.every(row => row[i] <= row[j])) {
                dp[i] = Math.max(dp[i], 1 + dp[j]);
            }
        }
    }
 
    // Return result
    return l - Math.max(...dp);
}
 
// Driver Code
let arr = ["hello", "geeks"];
let arr2 = ["xyz", "lmn", "pqr"];
 
// Function call to print required answer
console.log(deleteColumns(arr));
console.log(deleteColumns(arr2))
 
// This code is contributed by lokeshpotta20.


Output

1
0


Last Updated : 16 Mar, 2023
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