Given a string s of lowercase alphabets and a number k, find the minimum value of the string after removal of k characters. The value of a string is defined as the sum of squares of the count of each distinct character.
Examples:
Input: s = "abbccc", k = 2
Output: 6
Explanation: We remove two 'c' to get the value as 12 + 22 + 12 or We remove one 'b' and one 'c' to get the value 12 + 12 + 22.Input: s = "aaab", k = 2
Output: 2
Explanation: We remove two 'a'. Now we get the value as 12 + 12.
Try It Yourself
Table of Content
[Naive Approach] Frequency Count Reduction via Sorting
The main idea is to minimize the sum of squares of character frequencies by always reducing the highest frequency character first. Since higher frequencies contribute more to the square sum, decreasing them has the greatest impact. We count the frequency of each character, and for k steps, we reduce the current maximum frequency by 1, sorting after each reduction to maintain order. This balances the frequency distribution and lowers the overall square sum effectively.
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int minValue(string &s, int k) {
vector<int> freq(26, 0);
// Count frequency of each character
for (char ch : s) {
freq[ch - 'a']++;
}
// Perform k removals
while (k--) {
// Sort to bring the highest frequency at the end
sort(freq.begin(), freq.end());
// If the highest frequency is already 0, break early
if (freq[25] == 0) break;
// Decrease the highest frequency by 1
freq[25]--;
}
// Calculate the sum of squares
int result = 0;
for (int count : freq) {
result += count * count;
}
return result;
}
int main() {
string s = "abbccc";
int k = 2;
cout << minValue(s, k) << endl;
return 0;
}
import java.util.Arrays;
class GfG{
static int minValue(String s, int k) {
int[] freq = new int[26];
// Count frequency of each character
for (char c : s.toCharArray()) {
freq[c - 'a']++;
}
// Reduce highest frequency character k times
while (k > 0) {
// Sort so highest frequency is at the end
Arrays.sort(freq);
// No more characters to reduce
if (freq[25] == 0) break;
// Reduce the highest frequency by 1
freq[25]--;
k--;
}
// Calculate sum of squares of remaining frequencies
int result = 0;
for (int f : freq) {
result += f * f;
}
return result;
}
public static void main(String[] args) {
String s = "abbccc";
int k = 2;
System.out.println(minValue(s, k));
}
}
def minValue(s, k):
freq = [0] * 26
# Count frequency of each character
for c in s:
freq[ord(c) - ord('a')] += 1
# Reduce the highest frequency character k times
while k > 0:
# Sort so the max frequency is at the end
freq.sort()
# No characters left to reduce
if freq[25] == 0:
break
# Reduce the max frequency
freq[25] -= 1
k -= 1
# Calculate sum of squares of frequencies
result = sum(f * f for f in freq)
return result
if __name__ == "__main__":
s = "abbccc"
k = 2
print(minValue(s, k))
using System;
class GfG{
public static int minValue(string s, int k){
int[] freq = new int[26];
// Count frequency of each character
foreach (char c in s){
freq[c - 'a']++;
}
// Reduce highest frequency character k times
while (k > 0){
// Sort so max frequency is at the end
Array.Sort(freq);
// No characters left to reduce
if (freq[25] == 0) break;
// Reduce the max frequency
freq[25]--;
k--;
}
// Calculate sum of squares of frequencies
int result = 0;
foreach (int f in freq){
result += f * f;
}
return result;
}
// Main method for testing
public static void Main(){
string s = "abbccc";
int k = 2;
Console.WriteLine(minValue(s, k));
}
}
function minValue(s, k) {
const freq = Array(26).fill(0);
// Count frequency of each character
for (let i = 0; i < s.length; i++) {
freq[s.charCodeAt(i) - 'a'.charCodeAt(0)]++;
}
// Reduce the highest frequency character k times
while (k > 0) {
// Sort ascending
freq.sort((a, b) => a - b);
// No character to reduce
if (freq[25] === 0) break;
// Decrease highest frequency
freq[25]--;
k--;
}
// Calculate sum of squares of frequencies
let result = 0;
for (let f of freq) {
result += f * f;
}
return result;
}
// Driver Code
let s = "abbccc";
let k = 2;
console.log(minValue(s, k));
Output
6
Time Complexity: O(n + k * 26 log 26), for finding the frequency it's need O(n) and for each of the k steps, sorting the frequency array of size 26 takes constant time.
Auxiliary Space: O(1), Uses a fixed-size array freq array of size 26, so space usage is constant.
[Better Approach] Using Priority Queue - O(n + k log (26)) Time and O(1) Space
The main idea behind the solution is to reduce the impact of the most frequent characters in the string to minimize the sum of the squares of their frequencies. Since higher frequencies contribute more to the total sum (because of squaring), the algorithm uses a greedy approach: it removes one occurrence at a time from the most frequent character k times. After each removal, it recalculates the total by summing the squares of the remaining frequencies. This efficiently ensures the minimized total value.
#include <iostream>
#include <string>
#include <vector>
#include <queue>
using namespace std;
int minValue(string &s, int k) {
int n = s.length();
// If k is greater than or equal to the string length, return 0
if (k >= n) return 0;
// Frequency array for characters 'a' to 'z'
const int ALPHABET_SIZE = 26;
vector<int> freq(ALPHABET_SIZE, 0);
for (char ch : s) {
freq[ch - 'a']++;
}
// Max heap to always remove the character with highest frequency
priority_queue<int> pq;
for (int f : freq) {
if (f > 0)
pq.push(f);
}
// Remove k characters from the most frequent characters
while (k-- && !pq.empty()) {
int top = pq.top();
pq.pop();
if (top > 1)
pq.push(top - 1);
}
// Calculate the sum of squares of remaining frequencies
int result = 0;
while (!pq.empty()) {
int f = pq.top();
pq.pop();
result += f * f;
}
return result;
}
int main() {
string s = "abbccc";
int k = 2;
cout << minValue(s, k) << endl;
return 0;
}
import java.util.PriorityQueue;
import java.util.Collections;
class GfG {
static int minValue(String s, int k) {
int n = s.length();
if (k >= n) return 0;
// Frequency array for characters 'a' to 'z'
int[] freq = new int[26];
for (char ch : s.toCharArray()) {
freq[ch - 'a']++;
}
// Max heap (priority queue in descending order)
PriorityQueue<Integer> pq =
new PriorityQueue<>(Collections.reverseOrder());
for (int f : freq) {
if (f > 0) pq.add(f);
}
// Remove k characters from the most frequent characters
while (k-- > 0 && !pq.isEmpty()) {
int top = pq.poll();
if (top > 1) {
pq.add(top - 1);
}
}
// Calculate the sum of squares
int result = 0;
while (!pq.isEmpty()) {
int f = pq.poll();
result += f * f;
}
return result;
}
public static void main(String[] args) {
String s = "abbccc";
int k = 2;
System.out.println(minValue(s, k));
}
}
import heapq
def minValue(s, k):
n = len(s)
if k >= n:
return 0
# Frequency of each character
freq = [0] * 26
for ch in s:
freq[ord(ch) - ord('a')] += 1
# Use a max heap (by negating values)
max_heap = [-f for f in freq if f > 0]
heapq.heapify(max_heap)
# Remove k characters
while k > 0 and max_heap:
top = heapq.heappop(max_heap)
# decrement the frequency
top += 1
if top != 0:
heapq.heappush(max_heap, top)
k -= 1
# Compute sum of squares
return sum(x * x for x in max_heap)
# Driver code
if __name__ == "__main__":
s = "abbccc"
k = 2
print(minValue(s, k))
using System;
using System.Collections.Generic;
class GfG{
static int minValue(string s, int k){
int n = s.Length;
// If k is greater than or equal to string length, return 0
if (k >= n)
return 0;
// Count frequency of each character (a-z)
int[] freq = new int[26];
foreach (char c in s)
freq[c - 'a']++;
// Add non-zero frequencies to a list
List<int> counts = new List<int>();
foreach (int f in freq)
if (f > 0)
counts.Add(f);
// Remove k characters by decreasing the max frequency each time
while (k > 0 && counts.Count > 0){
// Sort in descending order to get the max
// frequency at the front
counts.Sort();
counts.Reverse();
// Decrease the highest frequency
if (counts[0] > 0)
counts[0]--;
k--;
}
// Compute sum of squares of remaining frequencies
int result = 0;
foreach (int count in counts)
result += count * count;
return result;
}
// Driver code
static void Main(){
string s = "abbccc";
int k = 2;
Console.WriteLine(minValue(s, k));
}
}
function minValue(s, k) {
const n = s.length;
// If k >= length, all characters can be removed
if (k >= n) return 0;
// Count frequency of each lowercase letter
const freq = Array(26).fill(0);
for (let i = 0; i < n; i++) {
const index = s.charCodeAt(i) - 'a'.charCodeAt(0);
freq[index]++;
}
// Create an array of non-zero frequencies
const q = [];
for (let i = 0; i < 26; i++) {
if (freq[i] > 0) {
q.push(freq[i]);
}
}
// Remove k characters from the highest frequencies
while (k > 0) {
// Sort descending to bring max to front
q.sort((a, b) => b - a);
// Decrease the max frequency
if (q[0] > 0) {
q[0]--;
k--;
}
}
// Calculate sum of squares of remaining frequencies
let result = 0;
for (let count of q) {
result += count * count;
}
return result;
}
// Driver Code
let s = "abbccc";
let k = 2;
console.log(minValue(s, k));
Output
6
[Expected Approach] Bucket-Based Frequency Reduction - O(n) Time and O(n) Space
The idea is to prioritizing the most frequent ones. By tracking how many characters have each frequency, we reduce the highest frequencies first since they contribute the most to the total sum. Each removal either decreases all characters at a max frequency or partially reduces some of them. After all removals, we compute the sum of squares of the remaining frequencies. This greedy approach ensures the minimal possible total.
Illustration:
#include <iostream>
#include <string>
#include <algorithm>
#include <vector>
using namespace std;
int minValue(string &s, int k) {
int n = s.length();
int alphabetCount[26] = {0};
int maxi = 0;
// Count frequency of each character
for (char c : s) {
alphabetCount[c - 'a']++;
maxi = max(alphabetCount[c-'a'], maxi);
}
// frequency bucket
vector<int> freq(maxi+1, 0);
int maxFreq = 0;
// Fill frequency bucket
for (int i = 0; i < 26; i++) {
if (alphabetCount[i] > 0) {
freq[alphabetCount[i]]++;
maxFreq = max(maxFreq, alphabetCount[i]);
}
}
// Reduce frequencies using k removals
while (k > 0 && maxFreq > 0) {
int countAtMax = freq[maxFreq];
if (countAtMax <= k) {
k -= countAtMax;
freq[maxFreq - 1] += countAtMax;
freq[maxFreq] = 0;
maxFreq--;
} else {
freq[maxFreq] -= k;
freq[maxFreq - 1] += k;
k = 0;
}
}
int result = 0;
for (int i = 1; i <= maxi; i++) {
result += i * i * freq[i];
}
return result;
}
// Driver code
int main() {
string s = "abbccc";
int k = 2;
cout << minValue(s, k) << endl;
return 0;
}
import java.util.*;
class GfG{
public static int minValue(String s, int k) {
int n = s.length();
int[] alphabetCount = new int[26];
int maxi = 0;
// Count frequency of each character
for (char c : s.toCharArray()) {
alphabetCount[c - 'a']++;
maxi = Math.max(maxi, alphabetCount[c-'a']);
}
int[] freq = new int[maxi + 1];
int maxFreq = 0;
// Fill frequency bucket
for (int count : alphabetCount) {
if (count > 0) {
freq[count]++;
maxFreq = Math.max(maxFreq, count);
}
}
// Reduce frequencies using k removals
while (k > 0 && maxFreq > 0) {
int countAtMax = freq[maxFreq];
if (countAtMax <= k) {
k -= countAtMax;
freq[maxFreq - 1] += countAtMax;
freq[maxFreq] = 0;
maxFreq--;
} else {
freq[maxFreq] -= k;
freq[maxFreq - 1] += k;
k = 0;
}
}
int result = 0;
for (int i = 1; i <= maxi; i++) {
result += i * i * freq[i];
}
return result;
}
public static void main(String[] args) {
String s = "abbccc";
int k = 2;
System.out.println(minValue(s, k));
}
}
def minValue(s, k):
n = len(s)
# Count of each letter (a-z)
alphabetCount = [0] * 26
maxi = 0
# Count frequency of each character
for c in s:
alphabetCount[ord(c) - ord('a')] += 1
maxi = max(maxi, alphabetCount[ord(c) - ord('a')])
maxFreq = 0
# freq[i] = number of characters with frequency i
freq = [0] * (maxi + 1)
# Fill frequency bucket and track the maximum frequency
for count in alphabetCount:
if count > 0:
freq[count] += 1
maxFreq = max(maxFreq, count)
# Remove k characters by reducing higher frequencies
while k > 0 and maxFreq > 0:
count_at_max = freq[maxFreq]
if count_at_max <= k:
# Can remove all characters at this frequency
k -= count_at_max
freq[maxFreq - 1] += count_at_max
freq[maxFreq] = 0
maxFreq -= 1
else:
# Partially remove only k characters
freq[maxFreq] -= k
freq[maxFreq - 1] += k
k = 0
# Calculate the result: sum of (freq^2 * number of chars with that freq)
result = 0
for i in range(1, maxi+1):
result += i * i * freq[i]
return result
# Driver code
if __name__ == "__main__":
s = "abbccc"
k = 2
print(minValue(s, k))
using System;
class GfG{
static int MinValue(string s, int k) {
int n = s.Length;
int[] alphabetCount = new int[26];
int maxi = 0;
// Count frequency of each character
foreach (char c in s) {
alphabetCount[c - 'a']++;
maxi = Math.Max(maxi, alphabetCount[c - 'a']);
}
int[] freq = new int[maxi + 1];
int maxFreq = 0;
// Fill frequency bucket
foreach (int count in alphabetCount) {
if (count > 0) {
freq[count]++;
maxFreq = Math.Max(maxFreq, count);
}
}
// Reduce frequencies using k removals
while (k > 0 && maxFreq > 0) {
int countAtMax = freq[maxFreq];
if (countAtMax <= k) {
k -= countAtMax;
freq[maxFreq - 1] += countAtMax;
freq[maxFreq] = 0;
maxFreq--;
} else {
freq[maxFreq] -= k;
freq[maxFreq - 1] += k;
k = 0;
}
}
int result = 0;
for (int i = 1; i <= maxi; i++) {
result += i * i * freq[i];
}
return result;
}
static void Main() {
string s = "abbccc";
int k = 2;
Console.WriteLine(MinValue(s, k));
}
}
function minValue(s, k) {
const n = s.length;
// Frequency of 'a' to 'z'
const alphabetCount = Array(26).fill(0);
let maxi = 0;
// Count frequency of each character
for (let i = 0; i < n; i++) {
alphabetCount[s.charCodeAt(i) - 97]++;
maxi = Math.max(maxi, alphabetCount[s.charCodeAt(i) - 97]);
}
let maxFreq = 0;
// freq[i] = number of characters with frequency i
const freq = Array(maxi + 1).fill(0);
// Fill frequency bucket and find the max frequency
for (let i = 0; i < 26; i++) {
const count = alphabetCount[i];
if (count > 0) {
freq[count]++;
maxFreq = Math.max(maxFreq, count);
}
}
// Reduce highest frequencies first until k becomes 0
while (k > 0 && maxFreq > 0) {
const countAtMax = freq[maxFreq];
if (countAtMax <= k) {
// Remove one occurrence from each character at maxFreq
k -= countAtMax;
freq[maxFreq - 1] += countAtMax;
freq[maxFreq] = 0;
maxFreq--;
}
else {
// Only part of the characters can be reduced
freq[maxFreq] -= k;
freq[maxFreq - 1] += k;
k = 0;
}
}
// Compute the final sum of squares
let result = 0;
for (let i = 1; i <= maxi; i++) {
result += i * i * freq[i];
}
return result;
}
// Driver code
const s = "abbccc";
const k = 2;
console.log(minValue(s, k));
Output
6