Count of alphabets having ASCII value less than and greater than k
Last Updated :
08 Feb, 2024
Given a string, the task is to count the number of alphabets having ASCII values less than and greater than or equal to a given integer k.
Examples:
Input: str = “GeeksForGeeks”, k = 90
Output:3, 10
G, F, G have ascii values less than 90.
e, e, k, s, o, r, e, e, k, s have ASCII values greater than or equal to 90
Input: str = “geeksforgeeks”, k = 90
Output: 0, 13
Approach: Start traversing the string and check if the current character has an ASCII value less than k. If yes then increment the count. So, the Remaining characters will have ASCII values greater than or equal to k. So, print len_of_String – count for characters with ASCII values greater than or equal to k.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int CountCharacters(string str, int k)
{
int cnt = 0;
int len = str.length();
for (int i = 0; i < len; i++) {
if (str[i] < k)
cnt++;
}
return cnt;
}
int main()
{
string str = "GeeksForGeeks";
int k = 90;
int count = CountCharacters(str, k);
cout << "Characters with ASCII values"
" less than K are "
<< count;
cout << "\nCharacters with ASCII values"
" greater than or equal to K are "
<< str.length() - count;
return 0;
}
|
Java
import java.util.*;
class GFG {
static int CountCharacters(String str, int k)
{
int cnt = 0;
int len = str.length();
for (int i = 0; i < len; i++) {
if (((int)str.charAt(i)) < k)
cnt++;
}
return cnt;
}
public static void main(String args[])
{
String str = "GeeksForGeeks";
int k = 90;
int count = CountCharacters(str, k);
System.out.println("Characters with ASCII values less than K are "+count);
System.out.println("Characters with ASCII values greater than or equal to K are "+(str.length() - count));
}
}
|
Python3
def CountCharacters(str, k):
cnt = 0
l = len(str)
for i in range(l):
if (ord(str[i]) < k):
cnt += 1
return cnt
if __name__ == "__main__":
str = "GeeksForGeeks"
k = 90
count = CountCharacters(str, k)
print ("Characters with ASCII values",
"less than K are", count)
print ("Characters with ASCII values",
"greater than or equal to K are",
len(str) - count)
|
C#
using System;
class GFG {
static int CountCharacters(String str, int k)
{
int cnt = 0;
int len = str.Length;
for (int i = 0; i < len; i++)
{
if (((int)str[i]) < k)
cnt++;
}
return cnt;
}
public static void Main()
{
String str = "GeeksForGeeks";
int k = 90;
int count = CountCharacters(str, k);
Console.WriteLine("Characters with ASCII values" +
"less than K are " + count);
Console.WriteLine("Characters with ASCII values greater" +
"than or equal to K are "+(str.Length - count));
}
}
|
Javascript
<script>
function CountCharacters(str,k)
{
let cnt = 0;
let len = str.length;
for (let i = 0; i < len; i++) {
if (str[i].charCodeAt(0) < k)
cnt++;
}
return cnt;
}
let str = "GeeksForGeeks";
let k = 90;
let count = CountCharacters(str, k);
document.write("Characters with ASCII values less than K are "+count+"<br>");
document.write("Characters with ASCII values greater than or equal to K are "+(str.length - count));
</script>
|
PHP
<?php
function CountCharacters($str, $k)
{
$cnt = 0;
$len = strlen($str);
for ($i = 0; $i < $len; $i++)
{
if ($str[$i] < chr($k))
$cnt += 1;
}
return $cnt;
}
$str = "GeeksForGeeks";
$k = 90;
$count = CountCharacters($str, $k);
echo("Characters with ASCII values" .
" less than K are " . $count);
echo("\nCharacters with ASCII values" .
" greater than or equal to K are " .
(strlen($str) - $count));
?>
|
Output
Characters with ASCII values less than K are 3
Characters with ASCII values greater than or equal to K are 10
Complexity Analysis:
- Time Complexity: O(N), as we are using a loop to traverse N times so it will cost us O(N) time
- Auxiliary Space: O(1), as we are not using any extra space.
Approach 2: Dynamic Programming:
Here’s a step-by-step explanation of how the DP algorithm works:
- Initialization: We first initialize a 2D array dp of size len x 128 (where len is the length of the string and 128 is the number of possible ASCII values) to all zeros. We also fill in the base cases, which are the values of dp[0][j] for all j. dp[0][j] is 1 if the first character of the string has ASCII value less than or equal to j, and 0 otherwise.
- Computing the DP table: We iterate through the string str from left to right. For each character str[i] and each possible ASCII value j, we compute dp[i][j] using the recurrence relation:
- dp[i][j] = dp[i-1][j] + (str[i] <= j)
- The first term on the right-hand side represents the number of characters in str[0:i-1] whose ASCII value is less than or equal to j. The second term is 1 if str[i] has ASCII value less than or equal to j, and 0 otherwise. By adding these two terms, we get the number of characters in str[0:i] whose ASCII value is less than or equal to j.
- Final answer: The final answer is given by dp[len-1][k-1], which represents the number of characters in the entire string str whose ASCII value is less than or equal to k-1.
- Output: We output the count of characters whose ASCII value is less than k, which is dp[len-1][k-1]. We also output the count of characters whose ASCII value is greater than or equal to k, which is str.length() – dp[len-1][k-1].
Here is the code of the above DP approach:
C++
#include <bits/stdc++.h>
using namespace std;
int CountCharacters(string str, int k)
{
int len = str.length();
int dp[len][128] = {0};
for (int j = 0; j < 128; j++) {
dp[0][j] = (str[0] <= j);
}
for (int i = 1; i < len; i++) {
for (int j = 0; j < 128; j++) {
dp[i][j] = dp[i-1][j] + (str[i] <= j);
}
}
return dp[len-1][k-1];
}
int main()
{
string str = "GeeksForGeeks";
int k = 90;
int count = CountCharacters(str, k);
cout << "Characters with ASCII values"
" less than K are "
<< count;
cout << "\nCharacters with ASCII values"
" greater than or equal to K are "
<< str.length() - count;
return 0;
}
|
Java
import java.util.Arrays;
public class CharacterCount {
static int countCharacters(String str, int k)
{
int len = str.length();
int[][] dp
= new int[len]
[128];
for (int j = 0; j < 128; j++) {
dp[0][j] = (str.charAt(0) <= j) ? 1 : 0;
}
for (int i = 1; i < len; i++) {
for (int j = 0; j < 128; j++) {
dp[i][j] = dp[i - 1][j]
+ ((str.charAt(i) <= j) ? 1 : 0);
}
}
return dp[len - 1][k - 1];
}
public static void main(String[] args)
{
String str = "GeeksForGeeks";
int k = 90;
int count = countCharacters(str, k);
System.out.println(
"Characters with ASCII values less than K are "
+ count);
System.out.println(
"Characters with ASCII values greater than or equal to K are "
+ (str.length() - count));
}
}
|
Python
def count_characters(s, k):
n = len(s)
dp = [[0] * 128 for _ in range(n)]
for j in range(128):
dp[0][j] = int(ord(s[0]) <= j)
for i in range(1, n):
for j in range(128):
dp[i][j] = dp[i - 1][j] + int(ord(s[i]) <= j)
return dp[n - 1][k - 1]
if __name__ == "__main__":
s = "GeeksForGeeks"
k = 90
count = count_characters(s, k)
print("Characters with ASCII values less than K are", count)
print("Characters with ASCII values greater than or equal to K are", len(s) - count)
|
C#
using System;
namespace CharacterCount {
class Program {
static int CountCharacters(string str, int k)
{
int len = str.Length;
int[][] dp = new int[len][];
for (int i = 0; i < len; i++) {
dp[i] = new int[128];
}
for (int j = 0; j < 128; j++) {
dp[0][j] = (str[0] <= j) ? 1 : 0;
}
for (int i = 1; i < len; i++) {
for (int j = 0; j < 128; j++) {
dp[i][j] = dp[i - 1][j]
+ ((str[i] <= j) ? 1 : 0);
}
}
return dp[len - 1][k - 1];
}
static void Main(string[] args)
{
string str = "GeeksForGeeks";
int k = 90;
int count = CountCharacters(str, k);
Console.WriteLine(
"Characters with ASCII values less than K are "
+ count);
Console.WriteLine(
"Characters with ASCII values greater than or equal to K are "
+ (str.Length - count));
}
}
}
|
Javascript
function countCharacters(str, k) {
const len = str.length;
const dp = new Array(len).fill(0).map(() => new Array(128).fill(0));
for (let j = 0; j < 128; j++) {
dp[0][j] = str.charCodeAt(0) <= j ? 1 : 0;
}
for (let i = 1; i < len; i++) {
for (let j = 0; j < 128; j++) {
dp[i][j] = dp[i - 1][j] + (str.charCodeAt(i) <= j ? 1 : 0);
}
}
return dp[len - 1][k - 1];
}
const str = "GeeksForGeeks";
const k = 90;
const count = countCharacters(str, k);
console.log("Characters with ASCII values less than K are " + count);
console.log("Characters with ASCII values greater than or equal to K are " + (str.length - count));
|
Output
Characters with ASCII values less than K are 3
Characters with ASCII values greater than or equal to K are 10
Time complexity: O(len * 128) because we iterate through the string str once and compute dp[i][j] for all i and j.
Space complexity: is O(len * 128) because we use a 2D array of size len x 128 to store the DP table.
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