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Jan 22, 2010 · Now, sum of natural numbers from 1 to N, can be expressed as Nx(N+1)/2. In your case N=100. Subtract the sum of the array from Nx(N+1)/2, where N=100. That is the missing number. The empty slot can be detected during the iteration in which the sum is computed. // will be the sum of the numbers in the array.
1. You need to use 2 instead of 1 , IntStream.rangeClosed(2, 10) You can also find min and max elements in the array using summaryStatistics(), private static int getMissingNumber(int[] a) {. IntSummaryStatistics summaryStatistics = Arrays.stream(a).summaryStatistics();
For big sorted arrays of unique numbers, you can binary search the array for either the lowest or highest unused number. Cost=Log2N. Example: 65536 items can be searched in 16 loops since
Mar 19, 2010 · 14. Turn the range you want to check into a HashSet: public IEnumerable<int> FindMissing(IEnumerable<int> values) {. HashSet<int> myRange = new HashSet<int>(Enumerable.Range(0,10)); myRange.ExceptWith(values); return myRange; } Will return the values that aren't in values.
Dec 21, 2013 · If the missing number is the highest or lowest one, ie. either 1 or 10 in his list is missing, then the above algorithm doesn't work, but that's nitpicking. – Lasse V. Karlsen Commented Dec 21, 2013 at 11:33
Apr 15, 2016 · Written this code, would like to get better approach using any algorithm to find missing numbers from an sorted or unsorted array. If its an unsorted array, i would sort and execute the following.
Mar 26, 2016 · Given an array A[] of length n, find a "missing" number k such that: k is not in A; 0<=k<=n; I've seen similar questions asked where the A[] contains numbers 1 to n with one number missing, but in this question, A[] can contain any numbers. I need a solution in O(n) time. For example,
Feb 5, 2010 · how to find missing number,given: two arrays as input, and find a number that is present in first array but is missing in second array 2 Find missing element from one array comparing from other in single for loop
Sep 13, 2012 · Iterate the numbers for the first bit, and divide the array to two halves - the first half has this bit as 0, the other half has it as 1. (Use the swap() for partitioning the array). Note that one half has ceil(N/2) elements, and the other has floor(N/2) elements. Repeat the process for the smaller array, until you find the missing number.
Nov 17, 2013 · Below is the generic answer in java code for any number of missing numbers in a given array //assumes that there are no duplicates a = [1,2,3,4,5] b = [1,2,5] a-b=[3,4] public list<integer> find(int[] input){ int[] a= new int[] {1,2,3,4,5};//create a new array without missing numbers List<Integer> l = new ArrayList<Integer>();//list for missing ...
