Finding The Frequency Of Elements In A Sorted Array | FREE DSA Course in JAVA | Lecture 71

We have to write a program in Java to find the frequency of all the elements of the given array.
Frequency means the number of times an element occurs in an array.
For eg, consider the array 20,20,30,30,30,40 the output will be:-
20 2
30 3
40 1
What's the logic for the program question.
You can simply find the solution by taking the frequency of element 1 as 1 by default and checking for further elements if they are equal to the previous element.
If yes increase the frequency variable by 1.
If the condition breaks, set the frequency variable to 1 again and do the same process.
In this way, you will be able to find the frequency of the elements.
There's one condition that we have missed in this logic.
Check out the complete program to know it and the code to write the program.
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#java #array #arrayjava #dsa #dsacourse #courseofdsa #datastructure #coding #programming

[@PetPalsParadise_]

Sir please never stop uploading content like this 🙏🏻 thank you

[@KaranSharma-lm6eh]

Following code covers all cases : int i = 1; int freq = 1; for( i =1; i

[@Itsallgoodmanedits]

Aww man thanks for the code

[@gottapukirankishorenaidu9411]

nice one bro its working

[@saikatmaity441]

good logic and nice explanation

[@mehakarora3977]

Thanku sir😊

[@Rajkamal9Kamal908]

Beginner friendly ❤❤

[@ajithsunkari2139]

Thanku sir

[@amitjain5273]

Sir, we can do this by a 2-pointer approach. As you had taught in the previous video. public static void frequency(int arr[]) { int firstElementIndex = 0; int duplicateElementIndex = 1; while (duplicateElementIndex < arr.length) { if (arr[firstElementIndex] == arr[duplicateElementIndex]) { duplicateElementIndex++; } else { System.out.println(arr[firstElementIndex] + " - " + (duplicateElementIndex - firstElementIndex)); firstElementIndex = duplicateElementIndex; } } System.out.println(arr[firstElementIndex] + " - " + (duplicateElementIndex - firstElementIndex)); }

[@jasper5016]

It's a very good solution.

[@dasparijoy2610]

Hi I am Pranay, if we use "i

[@babithatalari]

We can't do that bro.....we get an error

[@KaranSharma-lm6eh]

when i=a.length, then while comparing a[i] == a[i-1] will return out of bounds exception as a[i] doesn't exist means index = length -1.

[@KaranSharma-lm6eh]

Code is giving ArrayIndexOutOfBoundsException for int a[] = {20,20,20,30,30,45,46,46};

[@KaranSharma-lm6eh]

Index 8 out of bounds for length 8

[@user-tu2un9yp5b]

Sir this logic not work when we give more then 8 elements in an array

[@KaranSharma-lm6eh]

Please improve your logic as it is not working for different data sets such as if all the elements are duplicate in an Array, last 2 elements are same, in that case i will be at 2 positions ahead due to being incremented twice, one inside the inner while-loop and then after Sys.ot in the outer-while loop. So when the flow goes to if() loop- it checks for a[i-1] != a[i-2] but a[i-1] will throw indexOutOfBounds exception.

[@pritamroy8792]

If we put only same element in all index,it is giving result but showing "Index out of bound"

[@KaranSharma-lm6eh]

yes, that is what i too have observed. Overall code won't work for different data sets. As i will be at 2 positions ahead and a[i-1] will return ArrayIndexOutOfBounds exception.

[@bibekamatya8588]

int freq =1; for(int i = 0; i

[@the.sriya.guptaa]

can't we do this with for loop? plzz reply

[@mahaksoni7015]

yes we can

[@itachi_terabapp]

@@mahaksoni7015 how

[@harshrajushire429]

@@itachi_terabapp int ar[] = {20,20,40,30,30,30}; int freq =1; int i=1; for ( i= 1; i < ar.length; i++) { if(ar[i-1] == ar[i]){ freq++; }else{ System.out.println(ar[i-1] + " "+freq); freq=1; } } System.out.println(ar[i-1] + " "+freq); o/p: 20 2 40 1 30 3 ☠

[@itachi_terabapp]

@@harshrajushire429 ma chutya hu ...bro ...

[@AnoopSingh-fx2qy]

@@harshrajushire429 if the last elemnt repeats multiple time {20,20,40,50,50,30,30,30};

[@GaneshGuturi]

In language: private func FrequencyOfElementsInSortedArray(){ let input:[Int] = [5, 5, 10, 10, 10, 20, 20] // given sorted array let lastIndex = input.count - 1 var result = [Int: Int]() //hash map to store the value and frequency result[input[0]] = 1 for i in 1...lastIndex{ if result.keys.contains(input[i]){ let val = result[input[i]] ?? 0 result[input[i]] = val + 1 }else{ result[input[i]] = 1 } } print("result = \(result)") //Output = [5: 2, 10: 3, 20: 2] }

[@ShivaniKumari-qw3er]

static void findfreq(int ar[ ]){ int prev=0,l=0,count=0; While( l

[@ShivaniKumari-qw3er]

One extra closing curly bracket is included by mistake

[@user-on6dq2rf4p]

int[] intArray = { 10, 10, 20, 20, 20, 30, 40 }; int freq = 1, count = 1; if (intArray.Length > 1) { for (int i = 1; i < intArray.Length; i++) { if (intArray[i] == intArray[i - 1]) { freq++; count++; } else { Console.WriteLine($"Element: {intArray[i - 1]} Frequency: {freq}"); freq = 1; count++; if (intArray.Length == count) Console.WriteLine($"Element: {intArray[i]} Frequency: {freq}"); } } } else Console.WriteLine($"Element: {intArray[0]} Frequency: {freq}");

[@SahanasAllinOneChannel]

int[] a = {1, 2, 2, 3, 3, 3, 4}; for(int i=0; i

[@user-on6dq2rf4p]

int[] intArray = { 10, 10, 20, 20, 20, 30, 40 }; Dictionary elemFrequancy = new Dictionary(); foreach (var element in intArray) { if (!elemFrequancy.ContainsKey(element)) elemFrequancy.Add(element, 1); else elemFrequancy[element]++; } foreach (var item in elemFrequancy) { Console.WriteLine($"Element: {item.Key} Frequency: {item.Value}"); }

[@jasper5016]

Your tutorials are really very helpful. Thanks Here is my solution - private static void findFrequency() { int[] arr = {5,5,7,10,10,21}; //{10,20,20,30,30,30,40}; int index = 0; int freq = 1; while(index < arr.length){ if(index < arr.length-1 && arr[index] == arr[index + 1]){ freq++; } else { System.out.println(arr[index] + " " + freq); freq = 1; //reset frequency } index++; } }

[@KaranSharma-lm6eh]

what about edge cases - if array has only 1 element or if last 2 elements are same ?

[@KaranSharma-lm6eh]

your logic will be applicable only for case - when last 2 elements are different and arr.length >1.