Count columns to be deleted to make each row sorted
Last Updated :
16 Mar, 2023
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)
{
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;
}
}
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;
}
signed main()
{
vector<string> v{ "hello" , "geeks" };
vector<string> x{ "xyz" , "lmn" , "pqr" };
solve(v);
solve(x);
}
|
Java
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);
}
}
|
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)
if __name__ = = "__main__" :
v = [ "hello" , "geeks" ]
x = [ "xyz" , "lmn" , "pqr" ]
solve(v)
solve(x)
|
C#
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);
|
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;
int deleteColumns(vector<string>& A)
{
int l = A[0].length();
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 l - *max_element(dp.begin(), dp.end());
}
int main()
{
vector<string> arr = { "hello" , "geeks" };
vector<string> arr2 = { "xyz" , "lmn" , "pqr" };
cout << deleteColumns(arr) << endl;
cout << deleteColumns(arr2) << endl;
return 0;
}
|
Java
import java.util.*;
class Main {
static int deleteColumns(List<String> A) {
int l = A.get( 0 ).length();
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 l - Arrays.stream(dp).max().getAsInt();
}
public static void main(String[] args) {
List<String> arr = Arrays.asList( "hello" , "geeks" );
List<String> arr2 = Arrays.asList( "xyz" , "lmn" , "pqr" );
System.out.println(deleteColumns(arr));
System.out.println(deleteColumns(arr2));
}
}
|
Python3
def deleteColumns(A):
l = len (A[ 0 ])
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 l - max (dp)
arr = [ "hello" , "geeks" ]
arr2 = [ "xyz" , "lmn" , "pqr" ]
print (deleteColumns(arr) + "<br>" )
print (deleteColumns(arr2))
|
C#
using System;
using System.Collections.Generic;
using System.Linq;
public class MainClass {
static int DeleteColumns(List< string > A) {
int l = A[0].Length;
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 l - dp.Max();
}
public static void Main() {
List< string > arr = new List< string >() { "hello" , "geeks" };
List< string > arr2 = new List< string >() { "xyz" , "lmn" , "pqr" };
Console.WriteLine(DeleteColumns(arr));
Console.WriteLine(DeleteColumns(arr2));
}
}
|
Javascript
function deleteColumns(A) {
let l = A[0].length;
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 l - Math.max(...dp);
}
let arr = [ "hello" , "geeks" ];
let arr2 = [ "xyz" , "lmn" , "pqr" ];
console.log(deleteColumns(arr));
console.log(deleteColumns(arr2))
|
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