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Cognizant✅
Foundation assessment
9 OCT
Pattern :
2 Java codes
2 SQL queries
12 Mcq
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𝗕𝗼𝗼𝗸 𝘆𝗼𝘂𝗿 𝘀𝗹𝗼𝘁.👨💻
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Genc Foundation ✅
or any exam help available 🔥💯
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Cognizant Gen C learn
Exam Codes done✅
Full exam help
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Some basic Questions regarding the assessment
Question 1 - Can we give the assessment again on 30th Oct/ 6 nov, if we select for 9th October?
Answer - No, U can only attempt it once either 9th oct or 30th/6th nov.
Question 2 - What will be the pattern?
Answer - 2 Java Questions, 2 SQL queries, 15 MCQs ( from Basic web fundamentals n Java , SQL concepts)
Question 3 - What if you are not able to clear the assessment?
Answer - You will NOT BE revoked but your onboarding will be delayed & delay period depends upon business requirements so it can be of 1 month or upto 6+ months also.
Question 4 - When your CSD training will start?
Answer - CSD training will gets started within 2-3 weeks of your Pre - assessment results.
Question 5 - What will be assessment duration?
Answer - it will a 3 hours duration. Morning slot time starts from 10 AM - 1 PM.
Hope you got all your answers.
If u have anymore ,feel free to ask.
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Cognizant✅
Foundation assessment
Pattern :
2 Java codes
2 SQL queries
12 Mcq
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#include <iostream>
#include <vector>
#include <map>
#include <algorithm>
using namespace std;
char getCharWithKthHighestFreq(string s, int k) {
map<char, int> freq;
for (char c : s) {
freq[c]++;
}
map<int, vector<char>, greater<int>> grouped;
for (const auto &entry : freq) {
grouped[entry.second].push_back(entry.first);
}
int count = 0;
for (const auto &entry : grouped) {
if (count + entry.second.size() >= k) {
sort(entry.second.begin(), entry.second.end());
return entry.second[k - count - 1];
}
count += entry.second.size();
}
return -1; // If k is out of range
}
int main() {
int T;
cin >> T;
while (T--) {
string s;
int k;
cin >> s >> k;
char result = getCharWithKthHighestFreq(s, k);
cout << result << endl;
}
return 0;
}
"K FREQUENCY" AMAZON HACKON OA ROUND 2 CODE
VERIFIED 🥰
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#include <iostream>
#include <vector>
#include <cmath>
using namespace std;
bool isPrime(int n)
{
if (n <= 1)
return false;
for (int i = 2; i <= sqrt(n); i++)
if (n % i == 0)
return false;
return true;
}
int main() {
int n;
cin >> n;
vector<int> numbers(n);
for (int i = 0; i < n; i++) {
cin >> numbers[i];
}
int firstPrimeIndex = -1;
int lastPrimeIndex = -1;
for (int i = 0; i < n; i++) {
if (isPrime(numbers[i])) {
firstPrimeIndex = i;
break;
}
}
for (int i = n - 1; i >= 0; i--) {
if (isPrime(numbers[i])) {
lastPrimeIndex = i;
break;
}
}
if (firstPrimeIndex == -1) {
cout << -1;
} else {
cout << lastPrimeIndex - firstPrimeIndex;
}
return 0;
}
"Maximum Prime index" AMAZON HACKON OA ROUND 2 CODE
VERIFIED 🥰
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#include <iostream>
#include <vector>
#include <map>
using namespace std;
int main() {
int n;
cin >> n;
vector<int> numbers(n);
map<int, int> frequency;
for (int i = 0; i < n; i++) {
cin >> numbers[i];
frequency[numbers[i]]++;
}
bool isBalanced = true;
for (auto &entry : frequency) {
int a = entry.first;
int b = -entry.first;
if (frequency[a] != frequency[b]) {
isBalanced = false;
break;
}
}
if (isBalanced) {
cout << 1;
} else {
cout << 0;
}
return 0;
}
"Balance Array" AMAZON HACKON OA ROUND 2 CODE
VERIFIED 🥰
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#include <bits/stdc++.h>
using namespace std;
void countOccurrences() {
string inputString;
cin >> inputString;
int length = inputString.size();
vector<int> frequency(26, 0);
vector<int> result(length);
for (int i = length - 1; i >= 0; i--) {
result[i] = frequency[inputString[i] - 'a'];
frequency[inputString[i] - 'a']++;
}
// Using copy and ostream_iterator to print the elements
copy(result.begin(), result.end(), ostream_iterator<int>(cout, " "));
}
int main() {
countOccurrences();
return 0;
}
"countOccurrences" AMAZON HACKON OA ROUND 2 CODE
VERIFIED 🥰
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#include <iostream>
using namespace std;
long long find_minimum_X(long long A, long long B, long long C) {
long long X = 0;
for (int i = 31; i >= 0; i--) {
long long bitA = (A >> i) & 1;
long long bitB = (B >> i) & 1;
long long bitC = (C >> i) & 1;
if (bitC == 1) {
if (bitA == 0 && bitB == 0) {
X |= (1LL << i);
} else if (bitA == 1 && bitB == 0) {
return -1;
}
} else {
if (bitA == 1 && bitB == 1) {
X |= (1LL << i);
} else if (bitA == 0 && bitB == 1) {
return -1;
}
}
}
return X;
}
int main() {
long long A, B, C;
cin >> A >> B >> C;
long long result = find_minimum_X(A, B, C);
cout << result << endl;
return 0;
}
"BIT MANUPILATION" AMAZON HACKON OA ROUND 2 CODE
VERIFIED 🥰
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#include <iostream>
#include <vector>
using namespace std;
// Function to check if it's possible to transfer all diamonds in 'requiredTrips' trips with 'availableBoxes' boxes
bool canTransfer(const vector<int>& diamonds, int requiredTrips, int availableBoxes) {
int tripsNeeded = 0;
int currentBoxCapacity = 0;
for (int diamond : diamonds) {
if (diamond > availableBoxes) {
// If a single pouch contains more diamonds than the boxes available, return false.
return false;
}
if (currentBoxCapacity + diamond <= availableBoxes) {
// If we can add the diamonds from the current pouch to the current box, do so.
currentBoxCapacity += diamond;
} else {
// Otherwise, start a new trip with a new box.
tripsNeeded++;
currentBoxCapacity = diamond;
}
}
// Check if the trips needed are within the limit 'requiredTrips'.
return tripsNeeded < requiredTrips;
}
int main() {
int numberOfDiamonds, trips;
cin >> numberOfDiamonds; // Input size of the array
vector<int> diamondWeights(numberOfDiamonds);
for (int i = 0; i < numberOfDiamonds; i++) {
cin >> diamondWeights[i]; // Input the array elements
}
cin >> trips; // Input the number of trips
int minimumBoxes = 1; // Minimum possible boxes
int maximumBoxes = numberOfDiamonds; // Maximum possible boxes
while (minimumBoxes < maximumBoxes) {
int middleBoxes = minimumBoxes + (maximumBoxes - minimumBoxes) / 2;
if (canTransfer(diamondWeights, trips, middleBoxes)) {
// If it's possible to transfer all diamonds in 'trips' trips with 'middleBoxes' boxes,
// we can try to reduce the number of boxes.
maximumBoxes = middleBoxes;
} else {
// Otherwise, we need more boxes, so we increase the number of boxes to consider.
minimumBoxes = middleBoxes + 1;
}
}
// The 'minimumBoxes' variable now contains the minimum number of boxes required.
cout << minimumBoxes << endl;
return 0;
}
"DIAMOND BOX" AMAZON HACKON OA ROUND 2 CODE
VERIFIED 🥰
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#include <iostream>
int calculateCombinations(int totalSum, int maxNumber) {
int count = 0;
for (int i = 1; i <= maxNumber; i++) {
for (int j = 1; j <= maxNumber; j++) {
int thirdNumber = totalSum - i - j;
if (thirdNumber <= maxNumber && thirdNumber > 0) {
count++;
}
}
}
return count;
}
int main() {
int sum, maxNum;
cin >> sum >> maxNum;
cout << calculateCombinations(sum, maxNum);
return 0;
}
"TOY DISTRIBUTION" AMAZON HACKON OA ROUND 2 CODE
VERIFIED 🥰
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#include <iostream>
using namespace std;
int findSumOfDigits(int n) {
int sum = 0;
while (n > 0) {
sum += n % 10;
n /= 10;
}
return sum;
}
int main() {
int num;
cin >> num;
int originalNum = num;
while (num >= 10) {
num = findSumOfDigits(num);
}
cout << num * originalNum << endl;
return 0;
}
ELLA NUMBER GAME AMAZON HACKON OA ROUND 2 CODE
VERIFIED 🥰
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