Use the search below to find our solutions for selected questions!
Problem Statement Harold is a kidnapper who wrote a ransom note, but now he is worried it will be traced back to him through his handwriting. He found a magazine and wants to know if he can cut out whole words from it and use them to create an untraceable . . . Read more
Problem Statement You are given an unordered array consisting of consecutive integers without any duplicates. You are allowed to swap any two elements. You need to find the minimum number of swaps required to sort the array in ascending order. For example, given the array we perform the following steps: . . . Read more
Problem Statement Gary is an avid hiker. He tracks his hikes meticulously, paying close attention to small details like topography. During his last hike he took exactly steps. For every step he took, he noted if it was an uphill, , or a downhill, step. Gary’s hikes start and end . . . Read more
Problem Statement Create a stack data structure that supports the following operations as efficiently as possible: Push, which adds a new element atop the stack, Pop, which removes the top element of the stack, Find-Min, which returns (but does not remove) the smallest element of the stack Full Code
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package stacksqueues; import java.util.ArrayList; import java.util.LinkedList; import java.util.List; import datastructures.MyStack; /** * How would you design a stack which, in addition to push and pop, also has a function min which returns the minimum * element? Push, pop and min should all operate in O(1) time. * * @author lucas */ public class E2 { protected static class MyStackWithMin extends MyStack { MyStack mins; public MyStackWithMin() { mins = new MyStack<Integer>(); } public void push(int item) { super.push(item); if (item <= min()) { mins.push(item); } } public Integer pop() { int item = (int) super.pop(); if (item == min()) { mins.pop(); } return item; } /** * Return min element on stack * * @return */ public int min() { if (mins.isEmpty()) { return Integer.MAX_VALUE; } else { return (int)mins.peek(); } } } public static void main(String[] args) { MyStackWithMin stack = new MyStackWithMin(); stack.push(3); stack.push(31); stack.push(13); stack.push(2); stack.push(5); System.out.println(stack.min()); } } |
Problem statement Given a matrix in which each row and each column is sorted, write a method to find an element in it. Solution Assumptions: 1. Rows are sorted left to right in ascending order. Columns are sorted top to bottom in ascending order. 2. Matrix is of size . . . . Read more
Problem Statement Given an infinite number of quarters (25 cents), dimes (10 cents), nickels (5 cents) and pennies (1 cent), write code to calculate the number of ways of representing n cents Solution This is a recursive problem, so let’s figure out how to do makeChange(n) using prior solutions (i . . . Read more
Problem Statement Implement an algorithm to print all valid (e g , properly opened and closed) combinations of n-pairs of parentheses Example Input: 3 (e g , 3 pairs of parentheses) Output: ()()(), ()(()), (())(), ((())) Solution
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public class PrintAllValidParens { static void solve(int l, int r, int i, char[] str) { if (l == 0 && r == 0) { System.out.println(new String(str)); } else { if (l > 0) { str[i] = '('; solve(l - 1, r, i + 1, str); } if (r > 0 && r > l) { str[i] = ')'; solve(l, r - 1, i + 1, str); } } } public static void main(String[] args) { solve(3, 3, 0, new char[6]); } } |
Problem Statement Write a method to compute all permutations of a string Solution Let’s assume a given string represented by the letters To permute set , we can select the first character, , permute the remainder of the string to get a new list Then, with that new list, we . . . Read more
Problem statement You are given two 32-bit numbers, and , and two bit positions, and . Write a method to set all bits between and in equal to (e g , becomes a substring of located at and starting at ) Example Input: , , , Output: Solution
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public class SetBits { public static int updateBits(int n, int m, int i, int j) { int max = ~0; /* All 1’s */ // 1’s through position j, then 0’s int left = max - ((1 << j) - 1); System.out.println(Integer.toBinaryString(left)); // 1’s after position i int right = ((1 << i) - 1); System.out.println(Integer.toBinaryString(right)); // 1’s, with 0s between i and j int mask = left | right; // Clear i through j, then put m in there return (n & mask) | (m << i); } public static void main(String[] args) { updateBits(1024, 21, 2, 6); } } |
Problem Statement Given a sorted (increasing order) array, write an algorithm to create a binary tree with minimal height. Solution We will try to create a binary tree such that for each node, the number of nodes in the left subtree and the right subtree are equal, if possible Algorithm: . . . Read more