Programming Articles - Page 1123 of 3366

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

A matrix is Toeplitz if every diagonal from top-left to bottom-right has the same elements.Example 1[[1, 2, 3, 4], [5, 1, 2, 3], [9, 5, 1, 2]]Output −trueIn the above grid, the diagonals are −"[9]", "[5, 5]", "[1, 1, 1]", "[2, 2, 2]", "[3, 3]", "[4]".In each diagonal all elements are the same, so the answer is True.Example 2Input: matrix [[1, 2], [2, 2]]Output −falseThe diagonal "[1, 2]" has different elementsCodepublic class Matrix    {    public bool ToeplitzMatrix(int[, ] mat)    {       int row = getMatrixRowSize(mat);       int col = getMatrixColSize(mat);       ... Read More

The buffer object can be encoded and decoded into Base64 string. The buffer class can be used to encode a string into a series of bytes. The Buffer.from() method takes a string as an input and converts it into Base64.The converted bytes can be changed again into String. The toString() method is used for converting the Base64 buffer back into the string format.SyntaxBuffer.from(string, [encoding]) object.toString(encoding)ParametersThe parameters are described below:string − This input parameter takes input for the string that will be encoded into the base64 format.encoding − This input parameter takes input for the encoding in which string will be encoded and ... Read More