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anon.. would this work for a number like 17 (10001)?

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bool checkPalindrome(unsigned int n) { int m = n, k =0; bool ret = false; while(m!=0) { int i = 1; i = i & m; k = k > 1; } if((k^n)==0) { cout<<"Palindrome"< Menos

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You can access the list only 3 numbers at a time: It tells we can read three numbers one after other in N/3 Iterations. So Time complexity will be O(N/3) Menos

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This is sliding window problem and priority queue DS can be used to solve this.

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http://www.intel.com/software/products/documentation/vlin/mergedprojects/analyzer_ec/mergedprojects/reference_olh/mergedProjects/instructions/instruct32_hh/vc270.htm this might be helpful Menos

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unsigned int a=123; // Given number unsigned int b=0; /// temp variable int i; for(i=0; i>1; } Menos

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package array; import java.util.Hashtable; /* * You are given an array of numbers. * You need to print the length of the maximum continuous sequence that you encounter. * For example if input is [3,8,10,1,9,6,5,7,2 ], * the continuous sequences are{1,2,3} and {5,6,7,8,9,10} * the latter is the longest one so the answer becomes 6. * O(n) solution was asked for, * assuming you have a hash map which supports O(1) insertion and fetching operations * src: http://www.glassdoor.com/Interview/Microsoft-India-Interview-Questions-EI_IE1651.0,9_IL.10,15_IN115_IP6.htm */ public class FindLengthOfLongestRandomlyDistContinousSequenceOfNumber { public static void main(String[] str){ int a[] = {3,6,8,1,5,7,0,9,2,4,10,14,13,12,11}; System.out.println(findLength(a)); int b[] = {5,2,3,1,6,7,10,8}; System.out.println(findLength(b)); } private static int findLength(int[] a) { int result = 0; Hashtable h = new Hashtable(); for(int i=0; i result){ result = curr.maxValue; } if(next != null){ next = h.get( curr.max); next.maxValue = curr.maxValue; h.put(a[i]+1, next); next.min = curr.min; } if(prev != null){ prev = h.get(curr.min); prev.maxValue = curr.maxValue; h.put(a[i]-1, prev); prev.max = curr.max; } h.put(a[i], curr); } } return result; } } class Node{ int maxValue; int max; int min; public Node(int v, int m, int n){ maxValue = v; max = m; min = n; } } Menos

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static int findlongconsecutivesubseq(int[] arr) { HashSet s= new HashSet(); int ans=0; //put all entries in hashset for(int i=0;i ans) ans=j-arr[i]; } } return ans; } Menos

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@Ajit: What're the initial values of the max heap? What happens to the first value that comes in? Menos

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Use selection sort for 'max' ( or for 'min') If K > series.length/2 then reverse the criteria- like in stead of looking for 15th highest out of 20 element array - look for (20 -15 =) 5th lowest and so on.... Menos

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Google Hangout - Unable to do video chat - Check whether the camera is turned on - Check whether the connected device have camera option - Check whether the internet connection is switched off - Check the video chat after refreshing or restarting the app - Check whether the Gmail account had been logged out - Check whether the camera option is disabled in the device - Check whether the camera is not working in the device - Check by opening the hangout in the mobile or other device - Check by opening the hangout in different browser window - Check by opening the hangout in different browser - Check whether the javascript is disabled - Check after clearing the cache of the browser - Check after clearing the cookie of the browser Menos

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def maxRepeating(str): str = str.replace(" ", "") print(str) n = len(str) count = 0 res = str[0] cur_count = 1 # Traverse string except # last character for i in range(0, n-1): # If current character # matches with next if ( str[i] == str[i + 1]): cur_count += 1 # If doesn't match, update result # (if required) and reset count else: if cur_count > count: count = cur_count res = str[i] cur_count = 1 return res str = "Amazon is a great company as it haas AtoooZzz" print(maxRepeating(str)) Menos

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I can think of two ways: 1) traverse from both heads, get two length: long, short traverse again starting from Array1[long-short] and Array2[0], looking for the same pointer O(4n) time, O(1) space 2) use a hash table holds all seen pointers. traverse from both heads O(2n) time, O(n) space Menos

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Step 1: if the linked lists are not empty, find the length of each linked list. Since they have a common end point, we will get length of common area of the linked lists too. Step 2: find the difference between the length of both list. Start traverse the biggest list till the difference in length between them, provided one of them is bigger than the other. Step 3: now they are at the same remaining length. Traverse both of the together till both are pointing to the same next node.That node is the common node which is the merging point. If there is no such point, then they are not merged. Menos

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Hello, Can you please share how much time did the recruiter take to get to you with an offer? Menos

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General QA or a lot of coding related. Thanks

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No

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No not really because they are not allowed to do so.