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# Build Lowest Number by Removing n digits from a given number

• Difficulty Level : Hard
• Last Updated : 28 Jun, 2021

Given a string ‘str’ of digits and an integer ‘n’, build the lowest possible number by removing ‘n’ digits from the string and not changing the order of input digits.
Examples:

```Input: str = "4325043", n = 3
Output: "2043"

Input: str = "765028321", n = 5
Output: "0221"

Input: str = "121198", n = 2
Output: "1118"```

The idea is based on the fact that a character among first (n+1) characters must be there in resultant number. So we pick the smallest of first (n+1) digits and put it in result, and recur for the remaining characters. Below is complete algorithm.

```Initialize result as empty string
res = ""
buildLowestNumber(str, n, res)
1) If n == 0, then there is nothing to remove.
Append the whole 'str' to 'res' and return

2) Let 'len' be length of 'str'. If 'len' is smaller or equal
to n, then everything can be removed
Append nothing to 'res' and return

3) Find the smallest character among first (n+1) characters
of 'str'.  Let the index of smallest character be minIndex.
Append 'str[minIndex]' to 'res' and recur for substring after
minIndex and for n = n-minIndex

buildLowestNumber(str[minIndex+1..len-1], n-minIndex).```

Below is the implementation of the above algorithm:

## C++14

 `// C++ program to build the smallest number by removing n``// digits from a given number``#include ``using` `namespace` `std;`` ` `// A recursive function that removes 'n' characters from``// 'str' to store the smallest possible number in 'res'``void` `buildLowestNumberRec(string str, ``int` `n, string& res)``{``    ``// If there are 0 characters to remove from str,``    ``// append everything to result``    ``if` `(n == 0) {``        ``res.append(str);``        ``return``;``    ``}`` ` `    ``int` `len = str.length();`` ` `    ``// If there are more characters to remove than string``    ``// length, then append nothing to result``    ``if` `(len <= n)``        ``return``;`` ` `    ``// Find the smallest character among first (n+1)``    ``// characters of str.``    ``int` `minIndex = 0;``    ``for` `(``int` `i = 1; i <= n; i++)``        ``if` `(str[i] < str[minIndex])``            ``minIndex = i;`` ` `    ``// Append the smallest character to result``    ``res.push_back(str[minIndex]);`` ` `    ``// substring starting from minIndex+1 to str.length()``    ``// - 1.``    ``string new_str``        ``= str.substr(minIndex + 1, len - minIndex);`` ` `    ``// Recur for the above substring and n equals to``    ``// n-minIndex``    ``buildLowestNumberRec(new_str, n - minIndex, res);``}`` ` `// A wrapper over buildLowestNumberRec()``string buildLowestNumber(string str, ``int` `n)``{``    ``string res = ``""``;`` ` `    ``// Note that result is passed by reference``    ``buildLowestNumberRec(str, n, res);``   ` `    ``// Remove all the leading zeroes``    ``string ans = ``""``;``    ``int` `flag = 0;``    ``for` `(``int` `i = 0; i < res.length(); i++) ``    ``{``        ``if` `(res[i] != ``'0'` `|| flag == 1) ``        ``{``            ``flag = 1;``            ``ans += res[i];``        ``}``    ``}`` ` `    ``if` `(ans.length() == 0)``        ``return` `"0"``;``    ``else``        ``return` `ans;`` ` `    ` `}`` ` `// Driver program to test above function``int` `main()``{``    ``string str = ``"121198"``;``    ``int` `n = 2;``    ``cout << buildLowestNumber(str, n);``    ``return` `0;``}`

## Java

 `// Java program to build the smallest number``// by removing n digits from a given number``class` `GFG {``    ``static` `String res = ``""``;`` ` `    ``// A recursive function that removes``    ``// 'n' characters from 'str' to store``    ``// the smallest possible number in 'res'``    ``static` `void` `buildLowestNumberRec(String str, ``int` `n)``    ``{`` ` `        ``// If there are 0 characters to remove from str,``        ``// append everything to result``        ``if` `(n == ``0``) {``            ``res += str;``            ``return``;``        ``}`` ` `        ``int` `len = str.length();`` ` `        ``// If there are more characters to``        ``// remove than string length,``        ``// then append nothing to result``        ``if` `(len <= n)``            ``return``;`` ` `        ``// Find the smallest character among``        ``// first (n+1) characters of str.``        ``int` `minIndex = ``0``;``        ``for` `(``int` `i = ``1``; i <= n; i++)``            ``if` `(str.charAt(i) < str.charAt(minIndex))``                ``minIndex = i;`` ` `        ``// Append the smallest character to result``        ``res += str.charAt(minIndex);`` ` `        ``// substring starting from``        ``// minIndex+1 to str.length() - 1.``        ``String new_str = str.substring(minIndex + ``1``);`` ` `        ``// Recur for the above substring``        ``// and n equals to n-minIndex``        ``buildLowestNumberRec(new_str, n - minIndex);``    ``}`` ` `    ``static` `String lowestNumber(String str, ``int` `n)``    ``{``        ``buildLowestNumberRec(str, n);``        ``String ans = ``""``;``        ``int` `flag = ``0``;``        ``for` `(``int` `i = ``0``; i < res.length(); i++) {``            ``if` `(res.charAt(i) != ``'0'` `|| flag == ``1``) {``                ``ans += res.charAt(i);``            ``}``        ``}`` ` `        ``if` `(ans.length() == ``0``)``            ``return` `"0"``;``        ``else``            ``return` `ans;``    ``}`` ` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``String str = ``"390028"``;``        ``int` `n = ``2``;``        ``String ans = lowestNumber(str, n);`` ` `        ``System.out.println(ans);``    ``}``}`

## C#

 `// C# program to build the smallest number``// by removing n digits from a given number``using` `System;`` ` `class` `GFG {``    ``static` `String res = ``""``;`` ` `    ``// A recursive function that removes``    ``// 'n' characters from 'str' to store``    ``// the smallest possible number in 'res'``    ``static` `void` `buildLowestNumberRec(String str, ``int` `n)``    ``{`` ` `        ``// If there are 0 characters to remove from str,``        ``// append everything to result``        ``if` `(n == 0) {``            ``res += str;``            ``return``;``        ``}`` ` `        ``int` `len = str.Length;`` ` `        ``// If there are more characters to``        ``// remove than string length,``        ``// then append nothing to result``        ``if` `(len <= n)``            ``return``;`` ` `        ``// Find the smallest character among``        ``// first (n+1) characters of str.``        ``int` `minIndex = 0;``        ``for` `(``int` `i = 1; i <= n; i++)``            ``if` `(str[i] < str[minIndex])``                ``minIndex = i;`` ` `        ``// Append the smallest character to result``        ``res += str[minIndex];`` ` `        ``// substring starting from``        ``// minIndex+1 to str.length() - 1.``        ``String new_str = str.Substring(minIndex + 1);`` ` `        ``// Recur for the above substring``        ``// and n equals to n-minIndex``        ``buildLowestNumberRec(new_str, n - minIndex);``    ``}``   ` `    ``static` `String lowestNumber(String str, ``int` `n)``    ``{``         ``buildLowestNumberRec(str,n);``         ``String ans = ``""``;``         ``int` `flag = 0;``         ``for``(``int` `i = 0;i

Output:

`1118`

Below is an optimized code in C++ contributed by Gaurav Mamgain

## C++14

 `// C++ program to build the smallest number by removing``// n digits from a given number``#include ``using` `namespace` `std;`` ` `void` `insertInNonDecOrder(deque<``char``>& dq, ``char` `ch)``{`` ` `    ``// If container is empty , insert the current digit``    ``if` `(dq.empty())``        ``dq.push_back(ch);`` ` `    ``else` `{``        ``char` `temp = dq.back();`` ` `        ``// Keep removing digits larger than current digit``        ``// from the back side of deque``        ``while` `(temp > ch && !dq.empty()) {``            ``dq.pop_back();``            ``if` `(!dq.empty())``                ``temp = dq.back();``        ``}`` ` `        ``// Insert the current digit``        ``dq.push_back(ch);``    ``}``    ``return``;``}`` ` `string buildLowestNumber(string str, ``int` `n)``{``    ``int` `len = str.length();`` ` `    ``// Deleting n digits means we need to print k digits``    ``int` `k = len - n;`` ` `    ``deque<``char``> dq;``    ``string res = ``""``;`` ` `    ``// Leaving rightmost k-1 digits we need to choose``    ``// minimum digit from rest of the string and print it``    ``int` `i;``    ``for` `(i = 0; i <= len - k; i++)`` ` `        ``// Insert new digit from the back side in``        ``// appropriate position and/ keep removing``        ``// digits larger than current digit``        ``insertInNonDecOrder(dq, str[i]);`` ` `    ``// Now the minimum digit is at front of deque``    ``while` `(i < len) {`` ` `        ``// keep the minimum digit in output string``        ``res += dq.front();`` ` `        ``// remove minimum digit``        ``dq.pop_front();`` ` `        ``// Again insert new digit from the back``        ``// side in appropriate position and keep``        ``// removing digits larger than current digit``        ``insertInNonDecOrder(dq, str[i]);``        ``i++;``    ``}`` ` `    ``// Now only one element will be there in the deque``    ``res += dq.front();``    ``dq.pop_front();``    ``return` `res;``}`` ` `string lowestNumber(string str, ``int` `n)``{``    ``string res = buildLowestNumber(str, n);``     ` `    ``// Remove all the leading zeroes``    ``string ans = ``""``;``    ``int` `flag = 0;``    ``for` `(``int` `i = 0; i < res.length(); i++) {``        ``if` `(res[i] != ``'0'` `|| flag == 1) {``            ``flag = 1;``            ``ans += res[i];``        ``}``    ``}`` ` `    ``if` `(ans.length() == 0)``        ``return` `"0"``;``    ``else``        ``return` `ans;``}`` ` `// Driver program to test above function``int` `main()``{``    ``string str = ``"765028321"``;``    ``int` `n = 5;``    ``cout <
Output
`221`

Output:

`0221`

Time Complexity: O(N)
Space Complexity: O(N)

Approach:2

Let’s suppose the length of the given string num be n.so the result string will contain the length of n-k.

As we proceed to solve this problem we should make sure that the output string contains minimum values at their high weightage positions. so we ensure that by using a stack.

1. Return 0 if k >=n. and return num if k=0.
2. Create a stack and iterate through num string and push the value at that position if it is greater than the top element of the stack.
3. Iterate through the num string and if the integer value at that position is less than the top of the stack we will pop the stack and decrement k until we reach the condition where the top of the stack is less than the value we are looking at(while k>0) (by this we are making sure that most significant positions of the result are filled with minimum values).
4. If the k is still greater than 0 we will pop stack until k becomes 0.
5. Append the elements in the stack to the result string.
6. Delete leading zeroes from the result string.

Below is the implementation of the above approach:

## Java

 `// Java program for the above approach``import` `java.io.*;``import` `java.util.*;`` ` `class` `GFG {`` ` `    ``public` `static` `String removeKdigits(String num, ``int` `k)``    ``{``        ``StringBuilder result = ``new` `StringBuilder();``         ` `        ``// We have to delete all digits``        ``if` `(k >= num.length()) {``            ``return` `"0"``;``        ``}``        ``// Nothing to delete``        ``if` `(k == ``0``) {``            ``return` `num;``        ``}``        ``Stack s = ``new` `Stack();`` ` `        ``for` `(``int` `i = ``0``; i < num.length(); i++) {``            ``char` `c = num.charAt(i);``           ` `            ``// Removing all digits in stack that are greater``            ``// tha this digit(since they have higher``            ``// weightage)``            ``while` `(!s.isEmpty() && k > ``0` `&& s.peek() > c) {``                ``s.pop();``                ``k--;``            ``}``            ``// ignore pushing 0``            ``if` `(!s.isEmpty() || c != ``'0'``)``                ``s.push(c);``        ``}``       ` `        ``// If our k isnt 0 yet then we keep poping out the``        ``// stack until k becomes 0``        ``while` `(!s.isEmpty() && k > ``0``) {``            ``k--;``            ``s.pop();``        ``}``        ``if` `(s.isEmpty())``            ``return` `"0"``;``        ``while` `(!s.isEmpty()) {``            ``result.append(s.pop());``        ``}``        ``String str = result.reverse().toString();`` ` `        ``return` `str;``    ``}`` ` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``String s = ``"765028321"``;``        ``int` `k = ``5``;``        ``System.out.println(removeKdigits(s, ``5``));``    ``}``}``// this code is contributed by gireeshgudaparthi`

## C++

 `// C++ program for the above approach`` ` `#include ``using` `namespace` `std;`` ` `string removeKdigits(string num, ``int` `k)``{``    ``int` `n = num.size();``    ``stack<``char``> mystack;``    ``// Store the final string in stack``    ``for` `(``char` `c : num) {``        ``while` `(!mystack.empty() && k > 0``               ``&& mystack.top() > c) {``            ``mystack.pop();``            ``k -= 1;``        ``}`` ` `        ``if` `(!mystack.empty() || c != ``'0'``)``            ``mystack.push(c);``    ``}`` ` `    ``// Now remove the largest values from the top of the``    ``// stack``    ``while` `(!mystack.empty() && k--)``        ``mystack.pop();``    ``if` `(mystack.empty())``        ``return` `"0"``;`` ` `    ``// Now retrieve the number from stack into a string``    ``// (reusing num)``    ``while` `(!mystack.empty()) {``        ``num[n - 1] = mystack.top();``        ``mystack.pop();``        ``n -= 1;``    ``}``    ``return` `num.substr(n);``}`` ` `int` `main()``{``    ``string str = ``"765028321"``;``    ``int` `k = 5;``    ``cout << removeKdigits(str, k);``    ``return` `0;``}`
Output
`221`

Time complexity: O(N)
Space complexity: O(N)