Create lexicographically minimum String using given operation
Last Updated :
13 Apr, 2023
Given a string s of length N consisting of digits from 0 to 9. Find the lexicographically minimum sequence that can be obtained from given string s by performing the given operations:
- Selecting any position i in the string and delete the digit ‘d‘ at ith position,
- Insert min(d+1, 9) on any position in the string s.
Examples:
Input: s = ” 5217 “
Output: s = ” 1367 “
Explanation:
Choose 0th position, d = 5 and insert 6 i.e. min(d+1, 9) between 1 and 7 in string; s = ” 2167 “
Choose 0th position, d = 2 and insert 3 i.e. min(d+1, 9) between 1 and 6 in string; s = ” 1367 “
Input: s = ” 09412 “
Output: s = ” 01259 “
Explanation:
Choose 1st position, d = 9 and insert 9 i.e. min(d+1, 9) at last of string; s = ” 04129 “
Choose 1st position, d = 4 and insert 5 i.e. min(d+1, 9) between 2 and 9 in string; s = ” 0 1 2 5 9 “
Approach: Implement the idea below to solve the problem:
At each position i, we check if there is any digit sj such that sj < si and j > i. As we need the lexicographically minimum string, we need to bring the jth digit ahead of ith digit. So delete the ith digit and insert min(si+1, 9) behind the jth digit. Otherwise, if no lesser digits are present ahead of ith digit, keep the digit as it is and don’t perform the operation.
Follow the below steps to implement the above idea:
- Create a suffix vector to store the minimum digit in the right part of ith position in the string.
- Run a loop from N-2 to 0 and store the minimal digit found till index i from the end. [This can be done by storing suf[i] = min( suf[i+1], s[i] ) ].
- Create a result vector that will store the digits that will be present in the final lexicographically minimum string.
- Traverse for each position from i = 0 to N-1 in the string and check if there is any digit less than the current digit (d = s[i]-‘0’) on the right side:
- If yes then push min(d+1, 9) in result.
- Else push (d) in the result.
- Sort the vector result and print the values [Because we can easily arrange them in that way as there is no constraint on how many times we can perform the operation].
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
string minimumSequence(string& s)
{
int n = s.size();
vector<int> suf(n);
string sol = "";
suf[n - 1] = s[n - 1] - '0';
for (int i = n - 2; i >= 0; i--)
suf[i] = min(suf[i + 1], s[i] - '0');
vector<int> res;
for (int i = 0; i < n; i++) {
if (suf[i] < s[i] - '0')
res.push_back(min(9, s[i] - '0' + 1));
else
res.push_back(s[i] - '0');
}
sort(res.begin(), res.end());
for (int x : res)
sol += char(x + '0');
return sol;
}
int main()
{
string s = "09412";
cout << minimumSequence(s);
return 0;
}
|
Java
import java.io.*;
import java.util.*;
class GFG {
static String MinimumSequence(String s)
{
int n = s.length();
int[] suf = new int[n];
String sol = "";
suf[n - 1] = s.charAt(n - 1) - '0';
for (int i = n - 2; i >= 0; i--)
suf[i]
= Math.min(suf[i + 1], s.charAt(i) - '0');
int[] res = new int[n];
for (int i = 0; i < n; i++) {
if (suf[i] < s.charAt(i) - '0')
res[i] = Math.min(9, s.charAt(i) - '0' + 1);
else
res[i] = s.charAt(i) - '0';
}
Arrays.sort(res);
for (int i = 0; i < res.length; i++) {
sol += res[i];
}
return sol;
}
public static void main(String[] args)
{
String s = "09412";
System.out.println(MinimumSequence(s));
}
}
|
Python3
def minimumSequence(s):
n = len(s)
suf = [0 for _ in range(n)]
sol = ""
suf[n - 1] = ord(s[n - 1]) - ord('0')
for i in range(n-2, -1, -1):
suf[i] = min(suf[i + 1], ord(s[i]) - ord('0'))
res = []
for i in range(0, n):
if (suf[i] < ord(s[i]) - ord('0')):
res.append(min(9, ord(s[i]) - ord('0') + 1))
else:
res.append(ord(s[i]) - ord('0'))
res.sort()
for x in res:
sol += str(x)
return sol
if __name__ == "__main__":
s = "09412"
print(minimumSequence(s))
|
C#
using System;
using System.Linq;
public class GFG {
static string MinimumSequence(string s)
{
int n = s.Length;
int[] suf = new int[n];
string sol = "";
suf[n - 1] = s[n - 1] - '0';
for (int i = n - 2; i >= 0; i--)
suf[i] = Math.Min(suf[i + 1], s[i] - '0');
int[] res = new int[n];
for (int i = 0; i < n; i++) {
if (suf[i] < s[i] - '0')
res[i] = Math.Min(9, s[i] - '0' + 1);
else
res[i] = s[i] - '0';
}
Array.Sort(res);
for (int i=0;i<res.Length;i++){
sol += res[i].ToString();
}
return sol;
}
static void Main(string[] args)
{
string s = "09412";
Console.WriteLine(MinimumSequence(s));
}
}
|
Javascript
function minimumSequence(s)
{
let n = s.length;
let suf =new Array(n);
let sol = "";
suf[n - 1] = parseInt(s[n - 1]);
for (let i = n - 2; i >= 0; i--)
suf[i] = Math.min(suf[i + 1], parseInt(s[i]));
let res= [];
for (let i = 0; i < n; i++) {
if (suf[i] < parseInt(s[i]))
res.push(Math.min(9, parseInt(s[i]) + 1));
else
res.push(parseInt(s[i]));
}
res.sort();
for (let x=0 ; x<res.length; x++)
sol+=res[x];
return sol;
}
let s = "09412";
document.write(minimumSequence(s));
|
Time Complexity: O(N * log N)
Auxiliary Space: O(N)
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