DSA Interview asked questions 🫡#interviewquestions #datastructures #algorithms #dsa #jobs

[DarkoGaming]

Create a stack
Create a array
Put element one by one if (stack .contain(current elemnt )then add in array else add in stack


Return the array

[MohammedHasmi]

I think create another array and store frequency once loop is completed then count freq, which is easy probelm

[AmanSirENGINEER]

If we use Set so
Time complexity :- O(n)
Space complexity :- O(n)

[dakshbhardwaj]

a = [1, 2, 3, 4, 5, 6, 5, 7, 7, 8]
b = sorted(a)
c = []
for index, i in enumerate(b[:-1]):
if i == b[index + 1] and i not in c:
c.append(i)
if not c:
print(-1)
else:
print(c)

[tuljarammeghana]

We need to check if count > 1 for each element if it is greater then return the number

[travelbrain]

Create a vector for calculating requency asign each frequency as 0 after that make one more vector to store duplicates then run a loop to add frequency in the vector after that check with another loop if frequency is more than 1 pushback element in duplicate vector at last return it

[interestingduniya]

First find greatest element of array
create vector of that size
mark their count in that vector .
That create answer array and return
Tc 0(n) sc o(n)

[Manoj]

Sorting gives as O(n log n) right...how come, O(n) then..

[bhushanp]

Use for loop traversing compare current element with next element use count variable to keep track of each elements count check for the condition count >1 it will print the duplicate elements
Time complexity o(n)
Space complexity o(1)

[samarshelley]

Simply use frequency array and then iterate over array for finding duplicates, that's all. TC - O(N), SC-O(N).

[ashu.abhi]

we can do it by making frequency array

[Adithyan-zt]

sort the array and use temp var and store the value

[anshulthakur]

Use cycle sort for time O(N) and space O(1)

[SumanRout-vtrn]

No need to sort .
Use an auxiliary array for frequency of array elements

[ayushroy]

In java.
• Sort the Array
• Use a Stack
• Maintain an Array and Count
• Keep the Size of an array equal to count
• Check peek and the upcoming element if they are equal store it in an Array and update the count
• At the end after iterating the loop return the Array.

Time Complexity- O(n)
Space Complexity- O(n)

[League]

use frq ary and count 1 any difference than 1 will give duplicate

[laxmanbaisoya]

void checkDuplicate(int marks[], int size){
int temp = 0;
for(int i = 0; i<size; i++){
for(int j = i+1; j<size; j++){
if(marks[i] == marks[j]){
temp = marks[i];
printf("%d ", temp);
}
}
}
};

[rupamghosh]

Ek vector lelo jisme frequency store kardo
Baad me check katlo us frequency array me
Jis ka frequency== 2 hai wahi duplicate hai
Agr aisa kuch na mile toh return -1

[mansimehta]

Sorting use karke time complexity o(n logn) hogi

[Tundak_bhak]

Sort & then one loop very simple question that i have ever seen my life