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DSA Cprogramming~10 mins

Frequency Counter Pattern Using Hash Map in DSA C - Interactive Practice

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Practice - 5 Tasks
Answer the questions below
1fill in blank
easy

Complete the code to declare a hash map for counting frequencies.

DSA C
int [1][256] = {0};
Drag options to blanks, or click blank then click option'
Amap
Bcount
Cfreq
Dfrequency
Attempts:
3 left
💡 Hint
Common Mistakes
Using a variable name that is too long or unclear.
Not initializing the array to zero.
2fill in blank
medium

Complete the code to increase the frequency count of a character in the string.

DSA C
freq[(unsigned char)str[i]][1];
Drag options to blanks, or click blank then click option'
A++
B--
C+=
D=
Attempts:
3 left
💡 Hint
Common Mistakes
Using -- which decreases the count.
Using = which overwrites the count.
3fill in blank
hard

Fix the error in the condition to check if a character appears more than once.

DSA C
if (freq[(unsigned char)str[i]] [1] 1) {
Drag options to blanks, or click blank then click option'
A<=
B>
C<
D==
Attempts:
3 left
💡 Hint
Common Mistakes
Using == which only checks for exactly one occurrence.
Using <= or < which check for less or equal.
4fill in blank
hard

Fill both blanks to complete the loop that counts character frequencies in a string.

DSA C
for (int [1] = 0; [2] < length; [2]++) {
    freq[(unsigned char)str[i]]++;
}
Drag options to blanks, or click blank then click option'
Ai
Bj
Clength
Dcount
Attempts:
3 left
💡 Hint
Common Mistakes
Using different variables in the loop condition and increment.
Using a variable not declared in the loop.
5fill in blank
hard

Fill all three blanks to create a frequency counter that prints characters appearing more than once.

DSA C
for (int [1] = 0; [1] < 256; [1]++) {
    if (freq[[1]] [2] 1) {
        printf("%c appears %d times\n", (char)[1], freq[[1]]);
    }
}
Drag options to blanks, or click blank then click option'
Ai
B>
C==
D<
Attempts:
3 left
💡 Hint
Common Mistakes
Using different variables in the loop and inside the if condition.
Using the wrong comparison operator in the if condition.