Search in an almost sorted array; Find the closest pair from two sorted arrays; Find position of an element in a sorted array of infinite numbers; Find if there is a pair with a given sum in the rotated sorted Array; Kth largest element in a stream; Find the element that appears once in a sorted array We traverse given Binary Search Tree in reverse inorder and keep track of counts of nodes visited. Auxiliary Space: O(n) Find whether an array is subset of another array using Sorting and Design a stack that supports getMin() in O(1) time and O(1) extra space, Create a customized data structure which evaluates functions in O(1), Reverse a stack without using extra space in O(n), Check if a queue can be sorted into another queue using a stack, Count subarrays where second highest lie before highest, Delete array elements which are smaller than next or become smaller, Stack | Set 4 (Evaluation of Postfix Expression), Largest Rectangular Area in a Histogram using Stack, Find maximum of minimum for every window size in a given array, Expression contains redundant bracket or not, Check if a given array can represent Preorder Traversal of Binary Search Tree, Find maximum difference between nearest left and right smaller elements, Tracking current Maximum Element in a Stack, Range Queries for Longest Correct Bracket Subsequence Set | 2. Create a list to use values as index to store frequency of each element. Expected time complexity is O(n) and extra space is O(1). acknowledge that you have read and understood our, Data Structure & Algorithm Classes (Live), Full Stack Development with React & Node JS (Live), Preparation Package for Working Professional, Fundamentals of Java Collection Framework, Full Stack Development with React & Node JS(Live), GATE CS Original Papers and Official Keys, ISRO CS Original Papers and Official Keys, ISRO CS Syllabus for Scientist/Engineer Exam, Introduction to Stack Data Structure and Algorithm Tutorials, Applications, Advantages and Disadvantages of Stack, Design and Implement Special Stack Data Structure | Added Space Optimized Version, Design a stack with operations on middle element. ; If there is multiple highest element than the element which has lower value is This is also called the Median of Running Integers. Method 3 (Efficient): In the second method we simply calculate the second maximum element of the array and print all element which is less than or equal to the second maximum. 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Given an array arr[], find the maximum j i such that arr[j] > arr[i] Sliding Window Maximum (Maximum of all subarrays of size K) Sliding Window Maximum (Maximum of all subarrays of size k) using stack in O(n) time; Next Greater Element (NGE) for every element in given Array; Next greater element in same order as input; Next The Time Complexity of this solution is O(nLogn). Below is the step by step algorithm to solve this problem. Below is the implementation of the above problem. moves to the next element in that array. What is Array? A-143, 9th Floor, Sovereign Corporate Tower, We use cookies to ensure you have the best browsing experience on our website. After reading 10,median is 7.5 After reading 15 ,median is 10.Input: 1, 2, 3, 4Output: 1, 1.5, 2, 2.5Explanation: Given the input stream as an array of integers [1, 2, 3, 4]. For the first time it will add it to the map, next time it will replace the value. In the second traversal, find the smallest element greater than x. If a greater frequency element is found then that element is printed, otherwise -1 is printed. The inner loop looks for the first element whose frequency is greater than the frequency of the current element. Another way to find largest element by using library function. Max heap and min heap can be implemented using priority_queue in C++ STL. while traversing the array use these two pointers to place elements according to there range. For every element in the array update mp[array[i]]++ Store the element-frequency pair in a Priority Queue; Run a loop k times, and in each iteration remove the root of the priority queue and print the element. Input: arr[] = {1, 14, 5, 20, 4, 2, 54, 20, 87, 98, 3, 1, 32} , lowVal = 14 , highVal = 20Output: arr[] = {1, 5, 4, 2, 1, 3, 14, 20, 20, 98, 87, 32, 54}Explanation: all elements which are less than 14 are present in the range of 0 to 6 all elements which are greater than 14 and less than 20 are present in the range of 7 to 8 all elements which are greater than 20 are present in the range of 9 to 12, Input: arr[] = {1, 14, 5, 20, 4, 2, 54, 20, 87, 98, 3, 1, 32} , lowVal = 20 , highVal = 20 Output: arr[] = {1, 14, 5, 4, 2, 1, 3, 20, 20, 98, 87, 32, 54}. Time complexity: O(n)Auxiliary space: O(n). At last traverse the keys of map and print their respective values. ; Iterate over the remaining linked list. Iterate the array and save the element and its frequency in the hashmap. Read integers one by one and print the median correspondingly. Naive Approach: To solve the problem follow the below idea: Keep an array of size K. The idea is to keep the array sorted so that the K th largest element can be found in O(1) time (we just need to return the first element of the array, if the array is sorted in increasing order. Now use two loops. Calculate the new median as the average of top of elements of both max and min heap. For every pair, do a binary search for the second element in the given array, i.e., check if the second element of this pair exists as the first element in the array. For each node: Traverse the array from start to end. Write a program to reverse an array or string; Largest Sum Contiguous Subarray (Kadane's Algorithm) Scan the array and compute Maximum, second maximum and third maximum element present in the array. How to efficiently implement k stacks in a single array? If the frequency of the element which is pointed at the top of the stack is less than the frequency of the current element and the stack is not empty then pop. If the This will make sure that the bottom-most element for that horizontal distance is present on the map and if you see the tree from beneath that you will see that element. Calculate the new median as the average of top of elements of both max and min heap. Write an efficient program for printing k largest elements in an array. Data Structures & Algorithms- Self Paced Course, Complete Interview Preparation- Self Paced Course, Median of Stream of Running Integers using STL | Set 2, Median in a stream of integers (running integers), Finding Median of unsorted Array in linear time using C++ STL, Minimum product of k integers in an array of positive Integers, Randomized Algorithms | Set 3 (1/2 Approximate Median), Median of two sorted arrays of different sizes | Set 1 (Linear), C++ Program to Find median in row wise sorted matrix, Java Program to Find median in row wise sorted matrix. If yes, return the If there does not exist an answer for a position, then make the value -1. Using this method, we can overcome the problem of Method 1 which occurs when the smallest element is present in an array more than one time. One max heap to maintain elements of lower half and one min heap to maintain elements of higher half at any point of time.. For every newly read element, insert it into either max heap or min-heap and calculate the median based on the following conditions: If the size of max heap is greater than the size of min-heap and the element is less than the previous median then pop the top element from max heap and insert into min-heap and insert the new element to max heap else insert the new element to min-heap. O(mLog(m)) for sorting and O(nlog(m)) for binary searching each element of one array in another. Given an array with all distinct elements, find the largest three elements. 4) Do following for each element arr[i] a) Binary Search for the first occurrence of arr[i] + k in the sub array arr[i+1, N-1], let this index be X. In the first traversal find the minimum element. After the loop in step 3 is over, pop all the elements from stack and print -1 as next greater frequency element for them does not exist. Approach: Store the frequency of the elements of arr[] in a map say map1, with elements of arr[] as key and their frequency as value. In other words, we can get the median element as, when the input size is odd, we take the middle element of sorted data. All the elements in the array are distinct integers. Find the median of all the elements read so far starting from the first integer till the last integer. Mark the position of current element as i . Method 6(Using Binary Search)(Works with duplicates in the array): 1) Initialize count as 0. For every new element More than one such element can exist. "The holding will call into question many other regulations that protect consumers with respect to credit cards, bank accounts, mortgage loans, debt collection, credit reports, and identity theft," tweeted Chris Peterson, a former enforcement attorney at the CFPB who is If the input size is even, we pick an average of middle two elements in the sorted stream.Examples: Input: 5 10 15Output: 5, 7.5, 10Explanation: Given the input stream as an array of integers [5,10,15]. The individual elements of the three sets can appear in any order. Iterate the array and save the element and its frequency in the hashmap. How to process a new element of the stream? Kth Largest Element in BST when modification to BST is not allowed The second largest element is second last element in inorder traversal and second element in reverse inorder traversal. There is always a peak element. We keep track of two pointers, first to store next position of smaller element (smaller than range) from beginning, and second to store next position of greater element from end. Max heap and min heap can be implemented using priority_queue in C++ STL. We can see this property by creating some matrices using pen and paper. The idea is based on Dutch National Flag based QuickSort. So, after reading first element 1,median is 1. Kth Smallest/Largest Element in Unsorted Array | Set 1; Kth Smallest/Largest Element using STL; Kth smallest element in a row-wise and column-wise sorted 2D array | Set 1; K-th Largest Sum Contiguous Subarray; Program for Mean and median of an unsorted array; K maximum sums of overlapping contiguous sub-arrays Naive approach: Run nested loops and find the element in B[] which is not present in A[].The time complexity of this approach will be O(n 2). 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