Given a string S of Size N. Initially, the value of count is 0. The task is to find the value of count after N operations to remove all the N characters of the given string S where each operation is:
- In each operation, select alphabetically the smallest character in the string S and remove that character from S and add its index to count.
- If multiple characters are present then remove the character having the smallest index.
Note: Consider string as 1-based indexing.
Examples:
Input: N = 5, S = “abcab”
Output: 8
Explanation:
Remove character ‘a’ then string becomes “bcab” and the count = 1.
Remove character ‘a’ then string becomes “bcb” and the count = 1 + 3 = 4.
Remove character ‘b’ then string becomes “cb” and the count = 4 + 1 = 5.
Remove character ‘b’ then string becomes “c” and the count = 5 + 2 = 7.
Remove character ‘c’ then string becomes “” and the count = 7 + 1 = 8.Input: N = 5 S = “aabbc”
Output: 5
Explanation:
The value after 5 operations to remove all the 5 characters of String aabbc is 5.
Naive Approach: The idea is to check if the string is empty or not. If it is not empty then following are the steps to solve the problem:
- Find the first occurrence of the smallest alphabets in the current string, let’s say ind and remove it from the string.
- Increase the count by ind + 1.
- Repeat the above step until the string becomes empty.
Below is the implementation of the above approach:
// C++ program for the above approach #include <iostream> #include <string> using namespace std;
// Function to find the value after // N operations to remove all the N // characters of string S void charactersCount(string str, int n)
{ int count = 0;
// Iterate till N
while (n > 0) {
char cur = str[0];
int ind = 0;
for ( int j = 1; j < n; j++) {
if (str[j] < cur) {
cur = str[j];
ind = j;
}
}
// Remove character at ind and
// decrease n(size of string)
str.erase(str.begin() + ind);
n--;
// Increase count by ind+1
count += ind + 1;
}
cout << count << endl;
} // Driver Code int main()
{ // Given string str
string str = "aabbc" ;
int n = 5;
// Function call
charactersCount(str, n);
return 0;
} |
// Java program for the above approach import java.util.*;
class GFG{
// Function to find the value after // N operations to remove all the N // characters of String S static void charactersCount(String str, int n)
{ int count = 0 ;
// Iterate till N
while (n > 0 )
{
char cur = str.charAt( 0 );
int ind = 0 ;
for ( int j = 1 ; j < n; j++)
{
if (str.charAt(j) < cur)
{
cur = str.charAt(j);
ind = j;
}
}
// Remove character at ind and
// decrease n(size of String)
str = str.substring( 0 , ind) +
str.substring(ind + 1 );
n--;
// Increase count by ind+1
count += ind + 1 ;
}
System.out.print(count + "\n" );
} // Driver Code public static void main(String[] args)
{ // Given String str
String str = "aabbc" ;
int n = 5 ;
// Function call
charactersCount(str, n);
} } // This code is contributed by amal kumar choubey |
# Python3 program for the above approach # Function to find the value after # N operations to remove all the N # characters of String S def charactersCount( str , n):
count = 0 ;
# Iterate till N
while (n > 0 ):
cur = str [ 0 ];
ind = 0 ;
for j in range ( 1 , n):
if ( str [j] < cur):
cur = str [j];
ind = j;
# Remove character at ind and
# decrease n(size of String)
str = str [ 0 : ind] + str [ind + 1 :];
n - = 1 ;
# Increase count by ind+1
count + = ind + 1 ;
print (count);
# Driver Code if __name__ = = '__main__' :
# Given String str
str = "aabbc" ;
n = 5 ;
# Function call
charactersCount( str , n);
# This code is contributed by Amit Katiyar |
// C# program for the above approach using System;
class GFG{
// Function to find the value after // N operations to remove all the N // characters of String S static void charactersCount(String str, int n)
{ int count = 0;
// Iterate till N
while (n > 0)
{
char cur = str[0];
int ind = 0;
for ( int j = 1; j < n; j++)
{
if (str[j] < cur)
{
cur = str[j];
ind = j;
}
}
// Remove character at ind and
// decrease n(size of String)
str = str.Substring(0, ind) +
str.Substring(ind + 1);
n--;
// Increase count by ind+1
count += ind + 1;
}
Console.Write(count + "\n" );
} // Driver Code public static void Main(String[] args)
{ // Given String str
String str = "aabbc" ;
int n = 5;
// Function call
charactersCount(str, n);
} } // This code is contributed by gauravrajput1 |
<script> // Javascript program for the above approach
// Function to find the value after
// N operations to remove all the N
// characters of String S
function charactersCount(str, n)
{
let count = 0;
// Iterate till N
while (n > 0)
{
let cur = str[0].charCodeAt();
let ind = 0;
for (let j = 1; j < n; j++)
{
if (str[j].charCodeAt() < cur)
{
cur = str[j].charCodeAt();
ind = j;
}
}
// Remove character at ind and
// decrease n(size of String)
str = str.substring(0, ind) +
str.substring(ind + 1);
n--;
// Increase count by ind+1
count += ind + 1;
}
document.write(count + "</br>" );
}
// Given String str
let str = "aabbc" ;
let n = 5;
// Function call
charactersCount(str, n);
// This code is contributed by divyeshrabadiya07. </script> |
5
Time Complexity: O(N2)
Auxiliary Space: O(1)
Efficient Approach: This problem can be solved using the concept of Binary Index Tree or Fenwick Tree. Below are the steps:
- Initially, Store the values of indices of all the characters/alphabet in increasing order.
- Start with the smallest alphabet ‘a’ and remove all ‘a’s in the order of their occurrence. After removing, select the next alphabet ‘b’, and repeat this step until all alphabets are removed.
- Removing the character means that its corresponding value in the array is changed to 0, indicating that it is removed.
- Before removing, find the position of the character in the string using the sum() method in Fenwick Tree and add the position value to the count.
- After removing all the characters in the string, the value of count is obtained.
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Structure of Fenwick tree struct FenwickTree {
// Binary indexed tree
vector< int > bit;
// bit[0] is not involved
int n;
// Constructor
FenwickTree( int n)
{
this ->n = n + 1;
bit = vector< int >(n, 0);
}
// Constructor
FenwickTree(vector< int > a)
: FenwickTree(a.size())
{
for ( size_t i = 0; i < a.size(); i++)
add(i, a[i]);
}
// Sum of arr[0] + arr[1] + ...
// + arr[idx] where arr is array
int sum( int idx)
{
int ret = 0;
// Index in BITree[] is 1 more
// than the index in arr[]
idx = idx + 1;
// Traverse the ancestors of
// BITree[index]
while (idx > 0) {
// Add current element
// of BITree to sum
ret += bit[idx];
// Move index to parent
// node in getSum View
idx -= idx & (-idx);
}
// Return the result
return ret;
}
// Function for adding arr[idx]
// is equals to arr[idx] + val
void add( int idx, int val)
{
// Val is nothing but "DELTA"
// index in BITree[] is 1
// more than the index in arr[]
idx = idx + 1;
// Traverse all ancestors
// and add 'val'
while (idx <= n) {
// Add 'val' to current
// node of BI Tree
bit[idx] += val;
// Update index to that
// of parent in update View
idx += idx & (-idx);
}
}
// Update the index
void update( int idx, int val)
{
add(idx, val - valat(idx));
}
// Perform the sum arr[l] + arr[l+1]
// + ... + arr[r]
int sum( int l, int r)
{
return sum(r) - sum(l - 1);
}
int valat( int i)
{
return sum(i, i);
}
void clr( int sz)
{
bit.clear();
n = sz + 1;
bit.resize(n + 1, 0);
// Initially mark all
// are present (i.e. 1)
for ( int i = 0; i < n; i++)
add(i, 1);
}
}; // Function to count the characters void charactersCount(string str, int n)
{ int i, count = 0;
// Store the values of indexes
// for each alphabet
vector<vector< int > > mp
= vector<vector< int > >(26,
vector< int >());
// Create FenwickTree of size n
FenwickTree ft = FenwickTree(n);
ft.clr(n);
// Initially no indexes are
// stored for each character
for (i = 0; i < 26; i++)
mp[i].clear();
i = 0;
for ( char ch : str) {
// Push 'i' to mp[ch]
mp[ch - 'a' ].push_back(i);
i++;
}
// Start with smallest character
// and move towards larger character
// i.e., from 'a' to 'z' (0 to 25)
for (i = 0; i < 26; i++) {
// index(ind) of current character
// corres to i are obtained
// in increasing order
for ( int ind : mp[i]) {
// Find the number of characters
// currently present before
// ind using ft.sum(ind)
count += ft.sum(ind);
// Mark the corresponding
// ind as removed and
// make value at ind as 0
ft.update(ind, 0);
}
}
// Print the value of count
cout << count << endl;
} // Driver Code int main()
{ // Given string str
string str = "aabbc" ;
int n = 5;
// Function call
charactersCount(str, n);
return 0;
} |
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Structure of Fenwick tree struct FenwickTree { // Binary indexed tree
vector< int > bit;
// bit[0] is not involved
int n;
// Constructor
FenwickTree( int n)
{
this ->n = n + 1 ;
bit = vector< int >(n, 0 );
}
// Constructor
FenwickTree(vector< int > a)
: FenwickTree(a.size())
{
for (size_t i = 0 ; i < a.size(); i++)
add(i, a[i]);
}
// Sum of arr[0] + arr[1] + ...
// + arr[idx] where arr is array
int sum( int idx)
{
int ret = 0 ;
// Index in BITree[] is 1 more
// than the index in arr[]
idx = idx + 1 ;
// Traverse the ancestors of
// BITree[index]
while (idx > 0 ) {
// Add current element
// of BITree to sum
ret += bit[idx];
// Move index to parent
// node in getSum View
idx -= idx & (-idx);
}
// Return the result
return ret;
}
// Function for adding arr[idx]
// is equals to arr[idx] + val
void add( int idx, int val)
{
// Val is nothing but "DELTA"
// index in BITree[] is 1
// more than the index in arr[]
idx = idx + 1 ;
// Traverse all ancestors
// and add 'val'
while (idx <= n) {
// Add 'val' to current
// node of BI Tree
bit[idx] += val;
// Update index to that
// of parent in update View
idx += idx & (-idx);
}
}
// Update the index
void update( int idx, int val)
{
add(idx, val - valat(idx));
}
// Perform the sum arr[l] + arr[l+1]
// + ... + arr[r]
int sum( int l, int r)
{
return sum(r) - sum(l - 1 );
}
int valat( int i)
{
return sum(i, i);
}
void clr( int sz)
{
bit.clear();
n = sz + 1 ;
bit.resize(n + 1 , 0 );
// Initially mark all
// are present (i.e. 1)
for ( int i = 0 ; i < n; i++)
add(i, 1 );
}
}; // Function to count the characters void charactersCount(string str, int n)
{ int i, count = 0 ;
// Store the values of indexes
// for each alphabet
vector<vector< int > > mp
= vector<vector< int > >( 26 ,
vector< int >());
// Create FenwickTree of size n
FenwickTree ft = FenwickTree(n);
ft.clr(n);
// Initially no indexes are
// stored for each character
for (i = 0 ; i < 26 ; i++)
mp[i].clear();
i = 0 ;
for ( char ch : str) {
// Push 'i' to mp[ch]
mp[ch - 'a' ].push_back(i);
i++;
}
// Start with smallest character
// and move towards larger character
// i.e., from 'a' to 'z' (0 to 25)
for (i = 0 ; i < 26 ; i++) {
// index(ind) of current character
// corres to i are obtained
// in increasing order
for ( int ind : mp[i]) {
// Find the number of character
// currently present before
// ind using ft.sum(ind)
count += ft.sum(ind);
// Mark the corresponding
// ind as removed and
// make value at ind as 0
ft.update(ind, 0 );
}
}
// Print the value of count
cout << count << endl;
} // Driver Code int main()
{ // Given string str
string str = "aabbc" ;
int n = 5 ;
// Function call
charactersCount(str, n);
return 0 ;
} |
// C# program for the above approach using System;
using System.Collections.Generic;
class GFG{
// Structure of Fenwick tree public class FenwickTree
{ // Binary indexed tree
public int []bit;
// bit[0] is not involved
int n;
// Constructor
public FenwickTree( int n)
{
this .n = n + 1;
bit = new int [n];
}
// Constructor
public FenwickTree(List< int > a)
{
for ( int i = 0; i < a.Count; i++)
add(i, a[i]);
}
// Sum of arr[0] + arr[1] + ...
// + arr[idx] where arr is array
public int sum( int idx)
{
int ret = 0;
// Index in BITree[] is 1 more
// than the index in []arr
idx = idx + 1;
// Traverse the ancestors of
// BITree[index]
while (idx > 0)
{
// Add current element
// of BITree to sum
ret += bit[idx];
// Move index to parent
// node in getSum View
idx -= idx & (-idx);
}
// Return the result
return ret;
}
// Function for adding arr[idx]
// is equals to arr[idx] + val
public void add( int idx, int val)
{
// Val is nothing but "DELTA"
// index in BITree[] is 1
// more than the index in []arr
idx = idx + 1;
// Traverse all ancestors
// and add 'val'
while (idx <= n)
{
// Add 'val' to current
// node of BI Tree
bit[idx] += val;
// Update index to that
// of parent in update View
idx += idx & (-idx);
}
}
// Update the index
public void update( int idx, int val)
{
add(idx, val - valat(idx));
}
// Perform the sum arr[l] + arr[l+1]
// + ... + arr[r]
public int sum( int l, int r)
{
return sum(r) - sum(l - 1);
}
public int valat( int i)
{
return sum(i, i);
}
public void clr( int sz)
{
n = sz + 1;
bit = new int [n + 1];
// Initially mark all
// are present (i.e. 1)
for ( int i = 0; i < n; i++)
add(i, 1);
}
}; // Function to count the characters static void charactersCount(String str, int n)
{ int i, count = 0;
// Store the values of indexes
// for each alphabet
List< int > []mp = new List< int >[26];
for (i = 0; i < mp.Length; i++)
mp[i] = new List< int >();
// Create FenwickTree of size n
FenwickTree ft = new FenwickTree(n);
ft.clr(n);
// Initially no indexes are
// stored for each character
for (i = 0; i < 26; i++)
mp[i].Clear();
i = 0;
foreach ( char ch in str.ToCharArray())
{
// Push 'i' to mp[ch]
mp[ch - 'a' ].Add(i);
i++;
}
// Start with smallest character
// and move towards larger character
// i.e., from 'a' to 'z' (0 to 25)
for (i = 0; i < 26; i++)
{
// index(ind) of current character
// corres to i are obtained
// in increasing order
foreach ( int ind in mp[i])
{
// Find the number of characters
// currently present before
// ind using ft.sum(ind)
count += ft.sum(ind);
// Mark the corresponding
// ind as removed and
// make value at ind as 0
ft.update(ind, 0);
}
}
// Print the value of count
Console.Write(count + "\n" );
} // Driver Code public static void Main(String[] args)
{ // Given String str
String str = "aabbc" ;
int n = 5;
// Function call
charactersCount(str, n);
} } // This code is contributed by amal kumar choubey |
<script> // JavaScript program for the above approach // Structure of Fenwick tree class FenwickTree { constructor(n)
{
this .n = n + 1;
this .bit = new Array(n);
}
_FenwickTree(a)
{
for (let i = 0; i < a.length; i++)
this .add(i, a.get(i));
}
// Sum of arr[0] + arr[1] + ...
// + arr[idx] where arr is array
sum(idx)
{
let ret = 0;
// Index in BITree[] is 1 more
// than the index in arr[]
idx = idx + 1;
// Traverse the ancestors of
// BITree[index]
while (idx > 0)
{
// Add current element
// of BITree to sum
ret += this .bit[idx];
// Move index to parent
// node in getSum View
idx -= idx & (-idx);
}
// Return the result
return ret;
}
// Function for adding arr[idx]
// is equals to arr[idx] + val
add(idx,val)
{
// Val is nothing but "DELTA"
// index in BITree[] is 1
// more than the index in arr[]
idx = idx + 1;
// Traverse all ancestors
// and add 'val'
while (idx <= this .n)
{
// Add 'val' to current
// node of BI Tree
this .bit[idx] += val;
// Update index to that
// of parent in update View
idx += idx & (-idx);
}
}
// Update the index
update(idx,val)
{
this .add(idx, val - this .valat(idx));
}
// Perform the sum arr[l] + arr[l+1]
// + ... + arr[r]
_sum(l,r)
{
return this .sum(r) - this ._sum(l - 1);
}
valat(i)
{
return this .sum(i, i);
}
clr(sz)
{
this .n = sz + 1;
this .bit = new Array( this .n + 1);
for (let i=0;i< this .n+1;i++)
this .bit[i]=0;
// Initially mark all
// are present (i.e. 1)
for (let i = 0; i < n; i++)
this .add(i, 1);
}
} // Function to count the characters function charactersCount(str,n)
{ let i, count = 0;
// Store the values of indexes
// for each alphabet
let mp = new Array(26);
for (i = 0; i < mp.length; i++)
mp[i] = [];
// Create FenwickTree of size n
let ft = new FenwickTree(n);
ft.clr(n);
// Initially no indexes are
// stored for each character
for (i = 0; i < 26; i++)
mp[i]=[];
i = 0;
for (let ch of str.split( "" ).values())
{
// Push 'i' to mp[ch]
mp[ch.charCodeAt(0) - 'a' .charCodeAt(0)].push(i);
i++;
}
// Start with smallest character
// and move towards larger character
// i.e., from 'a' to 'z' (0 to 25)
for (i = 0; i < 26; i++)
{
// index(ind) of current character
// corres to i are obtained
// in increasing order
for (let ind of mp[i].values())
{
// Find the number of characters
// currently present before
// ind using ft.sum(ind)
count += ft.sum(ind);
// Mark the corresponding
// ind as removed and
// make value at ind as 0
ft.update(ind, 0);
}
}
// Print the value of count
document.write(count + "<br>" );
} // Driver Code // Given String str let str = "aabbc" ;
let n = 5; // Function call charactersCount(str, n); // This code is contributed by unknown2108 </script> |
# Python program for the above approach # Structure of Fenwick tree class FenwickTree:
# Constructor
def __init__( self , n):
self .n = n + 1
self .bit = [ 0 ] * self .n
# Constructor
def _FenwickTree( self , a):
for i in range ( len (a)):
self .add(i, a[i])
# Sum of arr[0] + arr[1] + ...
# + arr[idx] where arr is array
def sum ( self , idx):
ret = 0
# Index in BITree[] is 1 more
# than the index in arr[]
idx = idx + 1
# Traverse the ancestors of
# BITree[index]
while idx > 0 :
# Add current element
# of BITree to sum
ret + = self .bit[idx]
# Move index to parent
# node in getSum View
idx - = idx & ( - idx)
return ret
# Function for adding arr[idx]
# is equals to arr[idx] + val
def add( self , idx, val):
# Val is nothing but "DELTA"
# index in BITree[] is 1
# more than the index in arr[]
idx = idx + 1
# Traverse all ancestors
# and add 'val'
while idx < = self .n:
# Add 'val' to current
# node of BI Tree
self .bit[idx] + = val
# Update index to that
# of parent in update View
idx + = idx & ( - idx)
# Update the index
def update( self , idx, val):
self .add(idx, val - self .valat(idx))
# Perform the sum arr[l] + arr[l+1]
# + ... + arr[r]
def _sum( self , l, r):
return self . sum (r) - self . sum (l - 1 )
def valat( self , i):
return self . sum (i)
def clr( self , sz):
self .n = sz + 1
self .bit = [ 0 ] * ( self .n + 1 )
# Initially mark all
# are present (i.e. 1)
for i in range (sz):
self .add(i, 1 )
# Function to count the characters def charactersCount(s, n):
count = 0
# Store the values of indexes
# for each alphabet
mp = [[] for i in range ( 26 )]
# Create FenwickTree of size n
ft = FenwickTree(n)
ft.clr(n)
# Initially no indexes are
# stored for each character
for i, ch in enumerate (s):
mp[ ord (ch) - ord ( 'a' )].append(i)
# Start with smallest character
# and move towards larger character
# i.e., from 'a' to 'z' (0 to 25)
for i in range ( 26 ):
for ind in mp[i]:
# Find the number of characters
# currently present before
# ind using ft.sum(ind)
count + = ft. sum (ind)
# Mark the corresponding
# ind as removed and
# make value at ind as 0
ft.update(ind, 0 )
# Print the value of count
print (count)
# Driver Code s = "aabbc"
n = 5
charactersCount(s, n) # Contributed by adityasha4x71 |
5
Time Complexity: O(N*log(N))
Auxiliary Space: O(N)