weeklyupdate

C++/FindNumberswithEvenNumberofDigits.cpp

@@ -0,0 +1,23 @@
+/*
+
+Source: https://leetcode.com/problems/find-numbers-with-even-number-of-digits/
+
+Time: O(n), where n is the length of the array
+Space: O(1), in-place
+
+*/
+
+class Solution {
+public:
+ int findNumbers(vector<int>& nums) {
+
+ int count = 0;
+ for(int num : nums) {
+ if((num >= 10 && num < 100) || (num >= 1000 && num < 10000) || (num == 100000)) {
+ ++count;
+ }
+ }
+
+ return count;
+ }
+};

C++/KidsWiththeGreatestNumberofCandies.cpp

@@ -0,0 +1,30 @@
+/*
+
+Source: https://leetcode.com/problems/kids-with-the-greatest-number-of-candies/
+
+Time: O(N), where N is the length of candies array i.e. N is the number of elements present in the array
+Space: O(N), not in-place as we need a list of size equal to the size of candies array to store their corresponding boolean values
+
+*/
+
+class Solution {
+public:
+ vector<bool> kidsWithCandies(vector<int>& candies, int extraCandies) {
+
+ int max = 0;
+ for(int candy : candies) {
+ if(candy > max) {
+ max = candy;
+ }
+ }
+
+ max -= extraCandies;
+ vector<bool> res;
+
+ for(int candy : candies) {
+ res.push_back(candy >= max);
+ }
+
+ return res;
+ }
+};

C++/NumberofGoodPairs.cpp

@@ -0,0 +1,14 @@
+class Solution {
+public:
+ int numIdenticalPairs(vector<int>& nums) {
+
+ int freq[101] = {0};
+ int goodPairsCount = 0;
+
+ for(int num : nums) {
+ goodPairsCount += freq[num]++;
+ }
+
+ return goodPairsCount;
+ }
+};

C++/NumberofStepstoReduceaNumbertoZero.cpp

@@ -0,0 +1,26 @@
+/*
+
+Source: https://leetcode.com/problems/number-of-steps-to-reduce-a-number-to-zero/
+
+Time: O(logn), where log base2 n is the number of iterations done
+Space: O(1), in-place
+
+*/
+
+class Solution {
+public:
+ int numberOfSteps(int num) {
+
+ if(num == 0) {
+ return 0;
+ }
+
+ int steps = 0;
+ while(num != 0) {
+ steps += (num & 1) + 1;
+ num >>= 1;
+ }
+
+ return steps - 1;
+ }
+};

C++/SubtracttheProductandSumofDigitsofanInteger.cpp

@@ -0,0 +1,26 @@
+/*
+
+Source: https://leetcode.com/problems/subtract-the-product-and-sum-of-digits-of-an-integer/
+
+Time: O(n), where n is the given number
+Space: O(1), in-place
+
+*/
+
+class Solution {
+public:
+ int subtractProductAndSum(int n) {
+
+ int product = 1;
+ int sum = 0;
+
+ while(n != 0) {
+ int digit = n % 10;
+ product *= digit;
+ sum += digit;
+ n /= 10;
+ }
+
+ return product - sum;
+ }
+};

C++/UglyNumber.cpp

@@ -0,0 +1,32 @@
+/*
+
+Source: https://leetcode.com/problems/ugly-number/
+
+Time: O(log(base2)n), where log(base2)n is the max number of iteration done for a given integer n
+Space: O(1), in-place
+
+*/
+
+class Solution {
+public:
+ bool isUgly(int n) {
+
+ if(n <= 0) {
+ return false;
+ }
+
+ while((n & 1) == 0) {
+ n >>= 1;
+ }
+
+ while(n % 3 == 0) {
+ n /= 3;
+ }
+
+ while(n % 5 == 0) {
+ n /= 5;
+ }
+
+ return n == 1;
+ }
+};

C++/UglyNumberII.cpp

@@ -0,0 +1,47 @@
+/*
+
+Source: https://leetcode.com/problems/ugly-number-ii/
+
+Time: O(n), where n is a given positive integer
+Space: O(1), in-place
+
+*/
+
+class Solution {
+public:
+ int nthUglyNumber(int n) {
+
+ int uglyNumbers[n];
+ uglyNumbers[0] = 1;
+ int multipleOf2 = 2;
+ int multipleOf3 = 3;
+ int multipleOf5 = 5;
+ int ptr2 = 0;
+ int ptr3 = 0;
+ int ptr5 = 0;
+ int index = 1;
+
+ while(index < n) {
+
+ int minValue = (multipleOf2 < multipleOf3) ? multipleOf2 : multipleOf3;
+
+ if(multipleOf5 < minValue) {
+ minValue = multipleOf5;
+ }
+
+ uglyNumbers[index++] = minValue;
+
+ if(minValue == multipleOf2) {
+ multipleOf2 = uglyNumbers[++ptr2] << 1;
+ }
+ if(minValue == multipleOf3) {
+ multipleOf3 = 3 * uglyNumbers[++ptr3];
+ }
+ if(minValue == multipleOf5) {
+ multipleOf5 = 5 * uglyNumbers[++ptr5];
+ }
+ }
+
+ return uglyNumbers[index - 1];
+ }
+};

Java/FindNumberswithEvenNumberofDigits.java

@@ -0,0 +1,13 @@
+class Solution {
+ public int findNumbers(int[] nums) {
+
+ int count = 0;
+ for(int num : nums) {
+ if((num >= 10 && num < 100) || (num >= 1000 && num < 10000) || (num == 100000)) {
+ ++count;
+ }
+ }
+
+ return count;
+ }
+}

Java/KidsWiththeGreatestNumberofCandies.java

@@ -0,0 +1,28 @@
+/*
+
+Source: https://leetcode.com/problems/kids-with-the-greatest-number-of-candies/
+
+Time: O(N), where N is the length of candies array i.e. N is the number of elements present in the array
+Space: O(N), not in-place as we need a list of size equal to the size of candies array to store their corresponding boolean values
+
+*/
+
+class Solution {
+ public List<Boolean> kidsWithCandies(int[] candies, int extraCandies) {
+
+ int max = 0;
+ for(int candy : candies) {
+ if(candy > max) {
+ max = candy;
+ }
+ }
+
+ max -= extraCandies;
+ List<Boolean> res = new ArrayList<>(candies.length);
+ for(int candy : candies) {
+ res.add(candy >= max);
+ }
+
+ return res;
+ }
+}

Java/NumberofGoodPairs.java

@@ -0,0 +1,13 @@
+class Solution {
+ public int numIdenticalPairs(int[] nums) {
+
+ int[] freq = new int[101];
+ int goodPairsCount = 0;
+
+ for(int num : nums) {
+ goodPairsCount += freq[num]++;
+ }
+
+ return goodPairsCount;
+ }
+}

Java/NumberofStepstoReduceaNumbertoZero.java

@@ -0,0 +1,25 @@
+/*
+
+Source: https://leetcode.com/problems/number-of-steps-to-reduce-a-number-to-zero/
+
+Time: O(logn), where log base2 n is the number of iterations done
+Space: O(1), in-place
+
+*/
+
+class Solution {
+ public int numberOfSteps(int num) {
+
+ if(num == 0) {
+ return 0;
+ }
+
+ int steps = 0;
+ while(num != 0) {
+ steps += (num & 1) + 1;
+ num >>= 1;
+ }
+
+ return steps - 1;
+ }
+}

Java/SubtracttheProductandSumofDigitsofanInteger.java

@@ -0,0 +1,25 @@
+/*
+
+Source: https://leetcode.com/problems/subtract-the-product-and-sum-of-digits-of-an-integer/
+
+Time: O(n), where n is the given number
+Space: O(1), in-place
+
+*/
+
+class Solution {
+ public int subtractProductAndSum(int n) {
+
+ int product = 1;
+ int sum = 0;
+
+ while(n != 0) {
+ int digit = n % 10;
+ product *= digit;
+ sum += digit;
+ n /= 10;
+ }
+
+ return product - sum;
+ }
+}

Java/UglyNumber.java

@@ -0,0 +1,31 @@
+/*
+
+Source: https://leetcode.com/problems/ugly-number/
+
+Time: O(log(base2)n), where log(base2)n is the max number of iteration done for a given integer n
+Space: O(1), in-place
+
+*/
+
+class Solution {
+ public boolean isUgly(int n) {
+
+ if(n <= 0) {
+ return false;
+ }
+
+ while((n & 1) == 0) {
+ n >>= 1;
+ }
+
+ while(n % 3 == 0) {
+ n /= 3;
+ }
+
+ while(n % 5 == 0) {
+ n /= 5;
+ }
+
+ return n == 1;
+ }
+}

Java/UglyNumberII.java

@@ -0,0 +1,46 @@
+/*
+
+Source: https://leetcode.com/problems/ugly-number-ii/
+
+Time: O(n), where n is a given positive integer
+Space: O(1), in-place
+
+*/
+
+class Solution {
+ public int nthUglyNumber(int n) {
+
+ int[] uglyNumbers = new int[n];
+ uglyNumbers[0] = 1;
+ int multipleOf2 = 2;
+ int multipleOf3 = 3;
+ int multipleOf5 = 5;
+ int ptr2 = 0;
+ int ptr3 = 0;
+ int ptr5 = 0;
+ int index = 1;
+
+ while(index < n) {
+
+ int minValue = (multipleOf2 < multipleOf3) ? multipleOf2 : multipleOf3;
+
+ if(multipleOf5 < minValue) {
+ minValue = multipleOf5;
+ }
+
+ uglyNumbers[index++] = minValue;
+
+ if(minValue == multipleOf2) {
+ multipleOf2 = uglyNumbers[++ptr2] << 1;
+ }
+ if(minValue == multipleOf3) {
+ multipleOf3 = 3 * uglyNumbers[++ptr3];
+ }
+ if(minValue == multipleOf5) {
+ multipleOf5 = 5 * uglyNumbers[++ptr5];
+ }
+ }
+
+ return uglyNumbers[index - 1];
+ }
+}