Birthday Cake Candles
public static int birthdayCakeCandles(List<Integer> candles) {
// Write your code here
TreeMap<Integer,Integer> f = new TreeMap<>();
for (Integer a: candles)
f.put(a,f.getOrDefault(a, 0)+1);
return f.lastEntry().getValue();
}
https://www.hackerrank.com/challenges/mini-max-sum/problem
public static void miniMaxSum(List<Integer> arr) {
// Write your code here
Collections.sort(arr);
BigInteger sum = BigInteger.ZERO;
for (int i = 0; i < arr.size(); i++){
sum =sum.add(BigInteger.valueOf(arr.get(i)));
}
System.out.print(sum.subtract(BigInteger.valueOf(arr.get(arr.size()-1).intValue()))+" ");
System.out.print(sum.subtract(BigInteger.valueOf(arr.get(0).intValue())));
}
https://www.hackerrank.com/challenges/compare-the-triplets/problem
public static List<Integer> compareTriplets(List<Integer> a, List<Integer> b) {
// Write your code here
List<Integer> out = new ArrayList<Integer>(2);
out.add(0);
out.add(0);
for (int i = 0; i < a.size(); i++) {
if (a.get(i) > b.get(i))
out.set(0, out.get(0) + 1);
else if (a.get(i) < b.get(i))
out.set(1, out.get(1) + 1);
}
return out;
}Given an array of non-negative integers, count the number of unordered pairs of array
elements such that their bitwise AND is a power of 2.
For example, let’s say the array is arr = [10, 7, 2, 8, 3], and let ‘&’ denote the bitwise AND
operator. There are 6 unordered pairs of its elements that have a bitwise AND that is a
power of two:
For indices (0,1), 10 & 7 = 2, which is a power of 2.
For indices (0,2), 10 & 2 = 2, which is a power of 2.
For indices (0,3), 10 & 8 = 8, which is a power of 2.
For indices (0,4), 10 & 3 = 2, which is a power of 2.
For indices (1,2), 7 & 2 = 2, which is a power of 2.
For indices (2,4), 2 & 3 = 2, which is a power of 2.
Therefore, the answer is 6.
public static long countPairs(List<Integer> arr) {
// Write your code here
int count = 0;
for(int i = 0; i < arr.size(); i++){
for(int j= i+1; j < arr.size(); j++){
double a = Math.log(arr.get(i).intValue()&arr.get(j).intValue())/Math.log(2) ;
if(a == (int) a)
count++;
}
}
return count;
}
This is to certifya that
Omar GamalEldeeen
has successfully cleared the assessment for the skill SQL (Basic)
public static int equalizeArray(List<Integer> arr) {
HashMap<Integer, Integer> frequ = new HashMap<>();
for (Integer a : arr)
frequ.put(a, frequ.getOrDefault(a, 0) + 1);
int maxFrequ = Integer.MIN_VALUE;
for (Integer a : frequ.keySet()) {
maxFrequ = Integer.max(maxFrequ, frequ.get(a));
}
return arr.size() - maxFrequ;
}
class Sides {
int a;
int b;
int c;
Sides(int a, int b, int c) {
this.a = a;
this.b = b;
this.c = c;
}
}
public int compare(Sides o1, Sides o2) {
int[] o1Sides = new int[] { o1.a, o1.b, o1.c };
Arrays.sort(o1Sides);
int[] o2Sides = new int[] { o2.a, o2.b, o2.c };
Arrays.sort(o2Sides);
for (int i = 0; i < o1Sides.length; i++)
if (o1Sides[i] != o2Sides[i])
return -1;
return 0;
}
public int countDistinctTriangles(ArrayList<Sides> arr) {
if (arr.size() == 0)
return 0;
Collections.sort(arr, new Comparator<Sides>() {
@Override
public int compare(Sides o1, Sides o2) {
int[] o1Sides = new int[] { o1.a, o1.b, o1.c };
Arrays.sort(o1Sides);
int[] o2Sides = new int[] { o2.a, o2.b, o2.c };
Arrays.sort(o2Sides);
for (int i = 0; i < o1Sides.length; i++)
if (o1Sides[i] != o2Sides[i])
return -1;
return 0;
}
});
int count = 1;
for (int i = 1; i < arr.size(); i++) {
if (compare(arr.get(i), arr.get(i - 1)) != 0)
count++;
}
return count;
}
public boolean balancedSplitExists(int[] arr) {
Arrays.sort(arr);
int i = 0;
int j = arr.length - 1;
int leftSum = arr[0];
int rightSum = arr[arr.length-1];
while(i <= j) {
if(leftSum < rightSum) {
i++;
leftSum+=arr[i];
}else if (leftSum > rightSum) {
j--;
rightSum+=arr[j];
} else break;
}
if(j-i == 1) return true;
else return false;
}
public static int[] countSubarrays(int[] arr) {
int[] output = new int[arr.length];
for (int i = 0; i < arr.length; i++) {
int j = i;
int less = 0;
while(j>=0 && arr[j]<=arr[i]) {
less++;
j--;
}
output[i] = less;
j=i;
int greater = 0;
while(j < arr.length && arr[j] <= arr[i]) {
greater++;
j++;
}
output[i]+=greater-1;//one from duplication
}
return output;
}