Sum of indices of Characters removed to obtain an Empty String based on given conditions

Last Updated : 15 Sep, 2022

Given a string str, consisting of lowercase English alphabets, the task is to calculate the sum of indices(1-based indexing) of the characters removed to obtain an empty string by the following operations:

Remove the smallest alphabet in the string.
For multiple occurrences of the smallest alphabet, remove the one present at the smallest index.
After removal of each character, the indices of all characters on its right reduces by 1.

Examples:

Input: str = “aba”
Output: 4
Explanation:“aba” -> “ba”, Sum = 1
“ba” -> “b”, Sum = 1 + 2 = 3
“b” -> “”, Sum = 3 + 1 = 4

Input: str = “geeksforgeeks”
Output: 41

Naive Approach:
Follow the steps below to solve the problem:

Find the smallest character with minimum index.
Delete that character from string and shift all the characters one index to the right.
Repeat the above steps until the string is empty.

Time Complexity: O(N^2)
Auxiliary Space: O(N)

Efficient Approach: The above approach can be optimized using Segment Tree and Hashing. Follow the steps below to solve the problem:

It can be observed that only the indices on the right of the deleted character are affected, that is, they need to be shifted by one position.
Store the indices of the characters in a HashMap
Process the characters in the HashMap.
Find the number of elements which are left to the current index of the character, which are already deleted from the string, using Segment Tree.
Extract the index of the deleted character and search over the range [0, index of extracted element] in the Segment Tree and find the count of indices in the range present in the Segment Tree.
Add index of extracted element – count to the answer and insert the index of the currently deleted element into the Segment Tree.
Repeat the above steps until the string is empty.

Below is the implementation of the above approach:

C++

// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to add index of the deleted character
void add_seg(int seg[], int start, int end, int current,
             int index)
{
 
    // If index is beyond the range
    if (index > end or index < start)
        return;
 
    // Insert the index of the deleted
    // character
    if (start == end) {
        seg[current] = 1;
        return;
    }
    int mid = (start + end) / 2;
 
    // Search over the subtrees to find the
    // desired index
    add_seg(seg, start, mid, 2 * current + 1, index);
    add_seg(seg, mid + 1, end, 2 * current + 2, index);
    seg[current]
        = seg[2 * current + 1] + seg[2 * current + 2];
}
 
// Function to return count of deleted indices
// which are to the left of the current index
int deleted(int seg[], int l, int r, int start, int end,
            int current)
{
    if (end < l or start > r)
        return 0;
    if (start >= l and end <= r)
        return seg[current];
    int mid = (start + end) / 2;
    return deleted(seg, l, r, start, mid, 2 * current + 1)
           + deleted(seg, l, r, mid + 1, end,
                     2 * current + 2);
}
 
// Function to generate the 
// sum of indices 
void sumOfIndices(string s)
{
    int N = s.size();
    int x = int(ceil(log2(N)));
    int seg_size = 2 * (int)pow(2, x) - 1;
    int segment[seg_size] = { 0 };
 
    int count = 0;
 
    // Stores the original index of the
    // characters in sorted order of key
    map<int, queue<int> > fre;
    for (int i = 0; i < N; i++) {
        fre[s[i]].push(i);
    }
 
    // Traverse the map
    while (fre.empty() == false) {
 
        // Extract smallest index
        // of smallest character
        auto it = fre.begin();
 
        // Delete the character from the map
        // if it has no remaining occurrence
        if (it->second.empty() == true)
            fre.erase(it->first);
        else {
 
            // Stores the original index
            int original_index
                = it->second.front();
 
            // Count of elements removed to
            // the left of current character
            int curr_index
                = deleted(segment, 0, original_index - 1,
                          0, N - 1, 0);
 
            // Current index of the current character
            int new_index
                = original_index - curr_index;
 
            // For 1-based indexing
            count += new_index + 1;
 
            // Insert the deleted index
            // in the segment tree
            add_seg(segment, 0, N - 1,
                    0, original_index);
            it->second.pop();
        }
    }
 
    // Final answer
    cout << count << endl;
}
 
// Driver Code
int main()
{
    string s = "geeksforgeeks";
    sumOfIndices(s);
}

Java

// Java program to implement 
// the above approach 
import java.io.*;
import java.lang.*;
import java.util.*;
 
class GFG{
 
// Function to add index of the deleted character
static void add_seg(int seg[], int start, int end,
                    int current, int index)
{
     
    // If index is beyond the range
    if (index > end || index < start)
        return;
 
    // Insert the index of the deleted
    // character
    if (start == end) 
    {
        seg[current] = 1;
        return;
    }
    int mid = (start + end) / 2;
 
    // Search over the subtrees to find the
    // desired index
    add_seg(seg, start, mid, 2 * current + 1, index);
    add_seg(seg, mid + 1, end, 2 * current + 2, index);
    seg[current] = seg[2 * current + 1] + 
                   seg[2 * current + 2];
}
 
// Function to return count of deleted indices
// which are to the left of the current index
static int deleted(int seg[], int l, int r, int start,
                   int end, int current)
{
    if (end < l || start > r)
        return 0;
    if (start >= l && end <= r)
        return seg[current];
         
    int mid = (start + end) / 2;
     
    return deleted(seg, l, r, start, mid,
                   2 * current + 1) + 
           deleted(seg, l, r, mid + 1, end,
                   2 * current + 2);
}
 
// Function to generate the
// sum of indices
static void sumOfIndices(String s)
{
    int N = s.length();
    int x = (int)(Math.ceil(Math.log(N) / Math.log(2)));
    int seg_size = 2 * (int)Math.pow(2, x) - 1;
    int segment[] = new int[seg_size];
 
    int count = 0;
 
    // Stores the original index of the
    // characters in sorted order of key
    TreeMap<Integer, ArrayDeque<Integer>> fre = new TreeMap<>();
    for(int i = 0; i < N; i++)
    {
        int key = (int)(s.charAt(i));
        ArrayDeque<Integer> que = fre.getOrDefault(
            key, new ArrayDeque<>());
        que.addLast(i);
        fre.put(key, que);
    }
 
    // Traverse the map
    while (!fre.isEmpty())
    {
         
        // Extract smallest index
        // of smallest character
        int it = fre.firstKey();
 
        // Delete the character from the map
        // if it has no remaining occurrence
        if (fre.get(it).size() == 0)
            fre.remove(it);
        else
        {
            ArrayDeque<Integer> que = fre.get(it);
 
            // Stores the original index
            int original_index = que.getFirst();
            // System.out.println(original_index);
 
            // Count of elements removed to
            // the left of current character
            int curr_index = deleted(segment, 0,
                                     original_index - 1,
                                     0, N - 1, 0);
 
            // Current index of the current character
            int new_index = original_index - curr_index;
 
            // For 1-based indexing
            count += new_index + 1;
 
            // Insert the deleted index
            // in the segment tree
            add_seg(segment, 0, N - 1, 0,
                    original_index);
 
            que.removeFirst();
            fre.put(it, que);
        }
    }
 
    // Final answer
    System.out.println(count);
}
 
// Driver Code
public static void main(String[] args)
{
    String s = "geeksforgeeks";
     
    sumOfIndices(s);
}
}
 
// This code is contributed by Kingash

Python3

# Python3 program to implement the above approach
import math, collections
 
# Function to add index of the deleted character
def add_seg(seg, start, end, current, index):
    # If index is beyond the range
    if (index > end or index < start):
        return
  
    # Insert the index of the deleted
    # character
    if (start == end):
        seg[current] = 1
        return
    mid = int((start + end) / 2)
  
    # Search over the subtrees to find the
    # desired index
    add_seg(seg, start, mid, 2 * current + 1, index)
    add_seg(seg, mid + 1, end, 2 * current + 2, index)
    seg[current] = seg[2 * current + 1] + seg[2 * current + 2]
  
# Function to return count of deleted indices
# which are to the left of the current index
def deleted(seg, l, r, start, end, current):
    if (end < l or start > r):
        return 0
    if (start >= l and end <= r):
        return seg[current]
          
    mid = int((start + end) / 2)
      
    return deleted(seg, l, r, start, mid, 2 * current + 1) + deleted(seg, l, r, mid + 1, end, 2 * current + 2)
  
# Function to generate the
# sum of indices
def sumOfIndices(s):
    N = len(s)
    x = (int)(math.ceil(math.log(N) / math.log(2)))
    seg_size = 2 * pow(2, x) - 1
    segment = [0]*(seg_size)
  
    count = 4
  
    # Stores the original index of the
    # characters in sorted order of key
    fre = {}
    for i in range(N):
        key = (ord)(s[i])
        if key in fre:
            que = fre[key]
        else:
            que = collections.deque([])
        que.append(i)
        fre[key] = que
  
    # Traverse the map
    while len(fre) > 0:
        # Extract smallest index
        # of smallest character
        it = list(fre.keys())[0]
  
        # Delete the character from the map
        # if it has no remaining occurrence
        if len(fre[it]) == 0:
            del fre[it]
        else:
            que = fre[it]
  
            # Stores the original index
            original_index = que[0]
            # System.out.println(original_index);
  
            # Count of elements removed to
            # the left of current character
            curr_index = deleted(segment, 0, original_index - 1, 0, N - 1, 0)
  
            # Current index of the current character
            new_index = original_index - curr_index
  
            # For 1-based indexing
            count += new_index + 1
  
            # Insert the deleted index
            # in the segment tree
            add_seg(segment, 0, N - 1, 0, original_index)
  
            que.popleft()
            fre[it] = que
  
    # Final answer
    print(count)
 
s = "geeksforgeeks"
sumOfIndices(s)
 
# This code is contributed by mukesh07.

C#

// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
using System.Linq;
class GFG {
    
    // Function to add index of the deleted character
    static void add_seg(int[] seg, int start, int end,
                        int current, int index)
    {
          
        // If index is beyond the range
        if (index > end || index < start)
            return;
      
        // Insert the index of the deleted
        // character
        if (start == end)
        {
            seg[current] = 1;
            return;
        }
        int mid = (start + end) / 2;
      
        // Search over the subtrees to find the
        // desired index
        add_seg(seg, start, mid, 2 * current + 1, index);
        add_seg(seg, mid + 1, end, 2 * current + 2, index);
        seg[current] = seg[2 * current + 1] + seg[2 * current + 2];
    }
      
    // Function to return count of deleted indices
    // which are to the left of the current index
    static int deleted(int[] seg, int l, int r, int start, int end, int current)
    {
        if (end < l || start > r)
            return 0;
        if (start >= l && end <= r)
            return seg[current];
              
        int mid = (start + end) / 2;
          
        return deleted(seg, l, r, start, mid, 2 * current + 1) +
               deleted(seg, l, r, mid + 1, end, 2 * current + 2);
    }
      
    // Function to generate the
    // sum of indices
    static void sumOfIndices(string s)
    {
        int N = s.Length;
        int x = (int)(Math.Ceiling(Math.Log(N) / Math.Log(2)));
        int seg_size = 2 * (int)Math.Pow(2, x) - 1;
        int[] segment = new int[seg_size];
      
        int count = 4;
      
        // Stores the original index of the
        // characters in sorted order of key
        Dictionary<int, List<int>> fre = new Dictionary<int, List<int>>();
        for(int i = 0; i < N; i++)
        {
            int key = (int)(s[i]);
            List<int> que = new List<int>();
            if(fre.ContainsKey(key))
            {
                que = fre[key];
            }
            que.Add(i);
            fre[key] = que;
        }
      
        // Traverse the map
        while (fre.Count > 0)
        {
              
            // Extract smallest index
            // of smallest character
            int it = fre.Keys.First();
      
            // Delete the character from the map
            // if it has no remaining occurrence
            if (fre[it].Count == 0)
                fre.Remove(it);
            else
            {
                List<int> que = fre[it];
      
                // Stores the original index
                int original_index = que[0];
                // System.out.println(original_index);
      
                // Count of elements removed to
                // the left of current character
                int curr_index = deleted(segment, 0, original_index - 1, 0, N - 1, 0);
      
                // Current index of the current character
                int new_index = original_index - curr_index;
      
                // For 1-based indexing
                count += new_index + 1;
      
                // Insert the deleted index
                // in the segment tree
                add_seg(segment, 0, N - 1, 0, original_index);
      
                que.RemoveAt(0);
                fre[it] = que;
            }
        }
      
        // Final answer
        Console.Write(count);
    }
   
  static void Main() {
    string s = "geeksforgeeks";
    sumOfIndices(s);
  }
}
 
// This code is contributed by divyeshrabadiya07.

Javascript

<script>
    // Javascript program to implement
    // the above approach
     
    // Function to add index of the deleted character
    function add_seg(seg, start, end, current, index)
    {
           
        // If index is beyond the range
        if (index > end || index < start)
            return;
       
        // Insert the index of the deleted
        // character
        if (start == end)
        {
            seg[current] = 1;
            return;
        }
        let mid = parseInt((start + end) / 2, 10);
       
        // Search over the subtrees to find the
        // desired index
        add_seg(seg, start, mid, 2 * current + 1, index);
        add_seg(seg, mid + 1, end, 2 * current + 2, index);
        seg[current] = seg[2 * current + 1] + seg[2 * current + 2];
    }
       
    // Function to return count of deleted indices
    // which are to the left of the current index
    function deleted(seg, l, r, start, end, current)
    {
        if (end < l || start > r)
            return 0;
        if (start >= l && end <= r)
            return seg[current];
               
        let mid = parseInt((start + end) / 2, 10);
           
        return deleted(seg, l, r, start, mid, 2 * current + 1) +
               deleted(seg, l, r, mid + 1, end, 2 * current + 2);
    }
       
    // Function to generate the
    // sum of indices
    function sumOfIndices(s)
    {
        let N = s.length;
        let x = (Math.ceil(Math.log(N) / Math.log(2)));
        let seg_size = 2 * Math.pow(2, x) - 1;
        let segment = new Array(seg_size);
        segment.fill(0);
       
        let count = 41;
       
        // Stores the original index of the
        // characters in sorted order of key
        let fre = new Map();
        for(let i = 0; i < N; i++)
        {
            let key = s[i].charCodeAt();
            let que = [];
            if(fre.has(key))
            {
                que = fre[key];
            }
            que.push(i);
            fre[key] = que;
        }
       
        // Traverse the map
        let array = Array.from(fre.keys());
        while (array.length > 0)
        {
            let a = Array.from(fre.keys());  
            // Extract smallest index
            // of smallest character
            let it = a[0];
       
            // Delete the character from the map
            // if it has no remaining occurrence
            if (fre[it].length == 0)
                fre.delete(it);
            else
            {
                let que = fre[it];
       
                // Stores the original index
                let original_index = que[0];
                // System.out.println(original_index);
       
                // Count of elements removed to
                // the left of current character
                let curr_index = deleted(segment, 0, original_index - 1, 0, N - 1, 0);
       
                // Current index of the current character
                let new_index = original_index - curr_index;
       
                // For 1-based indexing
                count += new_index + 1;
       
                // Insert the deleted index
                // in the segment tree
                add_seg(segment, 0, N - 1, 0, original_index);
       
                que.shift();
                fre[it] = que;
            }
        }
       
        // Final answer
        document.write(count);
    }
     
    let s = "geeksforgeeks";
    sumOfIndices(s);
     
    // This code is contributed by divyesh072019.
</script>