array-algorithms

Given a sorted array of integers, how can we find a pair of integers that sum to K? e.g. array = 1,3,5,6,0 , K = 6 , the answer is 1 and 5. Time complexity should be minimized. You may want to look at this blog post: http://www.codingatwork.com/2011/07/array-sum/ My approach would be to do a binary search of the list for K/2 , then walk one variable a left and another variable b right trying to find a solution a+b=K . The idea would be to start a at the largest number less than or equal to K/2 and start b at the smallest number greater than a . Again, use a binary search to find a and let b be

It seems that the compiler is not going into the for loop.The sum of the array is to calculated.SumAll([1,4]) should return 10(1+2+3+4) as output. function sumAll(arr) { //return Math.max.apply(Math,arr); //return Math.min.apply(Math,arr); // return "0"; var sum=arr.reduce(function(a,b){ for(var i=Math.min.apply(Math,arr);i<=Math.max.apply(Math,arr);i++){ return a+b; } },0); //return sum; } sumAll([1, 4]); You could use directly the values from the array, without reduce. function sumAll(arr) { var i, sum = 0; for (i = Math.min.apply(null, arr); i <= Math.max.apply(null, arr); i++) { sum += i;

I am writing a code to find the maximum sum contiguous sub array in C. The logic seems fine according to me, but still the output is not correct. Please look into the code. The algorithm divides a bigger array into 2 sub-arrays. It then checks for maximum sum sub-array by examining the left array , right array and also the array containing the midpoint (It will check right and left from the midpoint and then return the maximum sum sub-array containing the midpoint). int* cross_max(int arr[], int low, int mid, int high) { int left_max, left_sum = -2000; int sum = 0; int i; for(i=mid; i>=low;i--

I want to sort an array and find the index of each element in the sorted order. So for instance if I run this on the array: [3,2,4] I'd get: [1,0,2] Is there an easy way to do this in Java? Louis Wasserman Let's assume your elements are stored in an array. final int[] arr = // elements you want List<Integer> indices = new ArrayList<Integer>(arr.length); for (int i = 0; i < arr.length; i++) { indices.add(i); } Comparator<Integer> comparator = new Comparator<Integer>() { public int compare(Integer i, Integer j) { return Integer.compare(arr[i], arr[j]); } } Collections.sort(indices, comparator);

I'm trying to solve the following exercise: Reverse an array without using the reverse method, without using a second array, and without duplicating any of the values. I've thought about making the array an object and then updating the array from the end to the beginning but I figured you can just update it as well. Tried something simple like: function reverseArray(array) { for (var i = 0; i < array.length; i++) { // var elem = array.shift(); var elem = array.shift() array.push(elem) } return array } array = ['a', 'b','c','d','e']; reverseArray(array); But that doesn't really change it. Any