Create a method MoveZeros, traverse through the array and count the number of Zeros in the array. Based on the count size make all the final cells to zero. Return without processing if the array length is null or empty. The final result will be in nums Array. Time complexity is O(N) because we are traversing through the array once.Time complexity − O(N)Space complexity − O(1)Examplepublic class Arrays{    public void MoveZeros(int[] nums){       if (nums == null || nums.Length == 0){          return;       }       int count = 0; ... Read More

Pascal’s Triangle is a number pattern in the shape of a triangle. Pascal’s Triangle has many applications in mathematics and statistics, including its ability to help you calculate combinations.Each number in the triangle is the sum of the two numbers above it. For example, row 4 − it’s the sum of 3 and 3 in the row above. The very first and very last number in any row are always going to be 1.Time complexity − O(N)Space complexity − O(N)Examplepublic class Arrays{    public List GeneratePascal(int n){       List res = new List();       if (n

Two Pointers pattern and is similar to quadruplet Sum to Zero. We can follow a similar approach to iterate through the array, taking one number at a time. At every step, we will save the difference between the quadruplet and the target number, and at each step we will compare it with the minimum target difference so far, so that in the end, we can return the triplet with the closest sum.Time complexitySorting the array will take O(N* logN). Overall fourSumClosest() will take O(N * logN + N^3), which is asymptotically equivalent to O(N^3).Space complexityThe space complexity of the above ... Read More

The easy approach is that we could create four nested loops and check one by one that the sum of all the four elements is zero or not. If the sum of the four elements is zero then print elements.Time Complexity − O(n4)Space Complexity − O(1)We could use an unordered set data structure to store each value of the array. Set provides the benefit of searching an element in O(1) time. So, for each pair in the array, we will look for the negative of their sum that might exist in the set. If such an element is found then ... Read More

Two Pointers pattern and is similar to Triplet Sum to Zero. We can follow a similar approach to iterate through the array, taking one number at a time. At every step, we will save the difference between the triplet and the target number, and at each step we will compare it with the minimum target difference so far, so that in the end, we can return the triplet with the closest sum.Time complexitySorting the array will take O(N* logN). Overall threeSumClosest() will take O(N * logN + N^2), which is asymptotically equivalent to O(N^2).Space complexityThe space complexity of the above ... Read More

Create 2 different arrays leftDis and rightDis. The leftDis will store the value when moved from left direction. The rightDis will store the shortest value when moved from right. Whenever the character is met add the position of the character to the array. In the last step calculate the maximum of both the arrays.Time Complexity − O(n)Space Complexity − O(n)Examplepublic class Arrays{    public int[] LongestDistanceToCharacter(string s, char c){       int stringLength = s.Length;       int[] leftDis = new int[s.Length];       int[] rightDis = new int[s.Length];       leftDis = Enumerable.Range(0, s.Length).Select(n => ... Read More

Create 2 different arrays leftDis and rightDis. The leftDis will store the value when moved from left direction. The rightDis will store the shortest value when moved from right. Whenever the character is met add the position of the character to the array. In the last step calculate the minimum of both the arrays.Time Complexity − O(n)Space Complexity − O(n)Examplepublic class Arrays{    public int[] ShortestDistanceToCharacter(string s, char c){       int stringLength = s.Length;       int[] leftDis = new int[s.Length];       int[] rightDis = new int[s.Length];       leftDis = Enumerable.Range(0, s.Length).Select(n => ... Read More

The easy approach is that we could create three nested loops and check one by one that the sum of all the three elements is zero or not. If the sum of the three elements is zero then print elements.Time Complexity − O(n3)Space Complexity − O(1)We could use an unordered set data structure to store each value of the array. Set provides the benefit of searching an element in O(1) time. So, for each pair in the array, we will look for the negative of their sum that might exist in the set. If such an element is found then ... Read More

The primitive solution for this problem is to scan all elements stored in the input matrix to search for the given key. This linear search approach costs O(MN) time if the size of the matrix is MxN.The matrix needs to be scanned from the top right, if the search element is greater than the top right element then increments the row or else decrement the column. The below piece of code develops a function SearchRowwiseIncrementedMatrix that takes a two-dimensional array and search key as input and returns either true or false depending upon the success or failure of search key ... Read More

The primitive solution for this problem is to scan all elements stored in the input matrix to search for the given key. This linear search approach costs O(MN) time if the size of the matrix is MxN.The matrix can be viewed as a sorted one-dimensional array. If all rows in the input matrix are concatenated in top-down order, it forms a sorted one-dimensional array. And, in that case binary search algorithm is suitable for this 2D array. The below piece of code develops a function SearchRowwiseColumnWiseMatrix that takes a two-dimensional array and search key as input and returns either true ... Read More