[ad_1] Deque: This would be a decent choice, since you can add/remove from both the front/end of the line in O(1) time, and the number of patients can be kept track of with an integer variable, so getting the size of the line would be O(1). You can also ensure a max capacity of N … Read more
[ad_1] I think (and you have confirmed) that you can represent your data as an XML structure. I this case, you can use XPath to find any branch in your data. Here is sample XML data inferred from your question: <?xml version=”1.0″ encoding=”utf-8″ ?> <stmt_echo> <subnodes> <exprs> <expr_constfetch> <subnodes> <name> <name> <subnodes> <parts> <part>true</part> <!– … Read more
[ad_1] $arr1 = [ [ ‘_date’ => ‘2019-10-16′,’_number’ => 1,’_order’ => 1, ‘name’ => ‘jack’,’other_ids’ => [‘b_id’ => 1253] ], [‘_date’ => 2020-10-11,’_number’ => 2,’_order’ => 2, ‘name’ => ‘joey’,’other_ids’ => [‘b_id’ => 1433] ] ]; $arr2 = [ [ ‘date’ => ‘2019-10-16′,’number’ => ‘1’,’order’ => ‘1’, ‘name’ => ‘jack’,’last_name’ => ‘foobar’,’other_ids’ => [‘b_id’ => … Read more
[ad_1] I got your problem. you need to declare the String[] hotel or String[] rooms as the global member. The scope of the String[] hotel would have been gone when the function execution is completed if you are using in another function or in main then it will be a different String array. So only … Read more
[ad_1] You are getting response in JSONObject and you are trying to get it in JSONArray.. thats why you are getting error.. Try this way… try { JSONObject result = new JSONObject(response); if(data.has(“ValidateLoginResult”){ JSONArray array = result.getJSONArray(“ValidateLoginResult”); for (int i = 0; i < array.length(); i++) { JSONObject obj = array.getJSONObject(i); String ErrorMessage= “”+obj.getString(“ErrorMessage”); String … Read more
[ad_1] #include <algorithm> //… bool isKeyword( const std::string &s ) { return ( std::find( keyword_list.begin(), keyword_list.end(), s ) != keyword_list.end() ); } If the vector would be sorted then you could use standard algorithm std::binary_search For example #include <algorithm> //… bool isKeyword( const std::string &s ) { return ( std::binary_search( keyword_list.begin(), keyword_list.end(), s ) ); … Read more
[ad_1] Dont do int characters [] [8]; , but string characters [] [8]; . The type you specify at the start of the array declaration is the type of info the array will store. Peace. 6 [ad_2] solved Placing Random String values into an int Array
[ad_1] const result = {}; result.items = [{ “organizationCode”: “FP1”, “organizationName”: “FTE Process Org” }, { “organizationCode”: “T11”, “organizationName”: “FTE Discrete Org” }, { “organizationCode”: “M1”, “organizationName”: “Seattle Manufacturing” } ]; let inputText = “starts with M”; // example input text const lastSpaceIndex = inputText.lastIndexOf(‘ ‘); const secondPartOfInput = inputText.substring(lastSpaceIndex + 1).trim(); const firstPartOfInput = … Read more
[ad_1] First get all data into an array. Then compose that array multidimensional array. So you can push two rows in every single key something like that. while($row = mysql_fetch_assoc($query)){ $data[] = $row; } $last_data = array(); $key = 0; for($i=0;$i<sizeof($data);$i++) { if($i%2 == 0) { $key++; } $last_data[$key][] = $data[$i]; } Then you can … Read more
[ad_1] arr[-1] gives you the last element of arr, arr[-2] the second-to-last element and so on. arr[2, 3] gives you from the element at index 2, three elements of arr arr[2..3] gives you from the element at index 2 to the element at index 3 2 [ad_2] solved Accessing ruby elements
[ad_1] Step 1: Combine all the arrays Step 2: Sort arrays using Arrays.sort(array); Step 3: Remove the duplicates. int[] a = {1,2,3,4,5}; int[] b = {1,2,3,4,5,6}; int[] c = {1,3,7}; int[] d = {2,3,4,8,9,10}; int[] resultArray1 = new int[a.length+b.length+c.length+d.length]; int arrayIndex = 0; for (int i=0; i< a.length ; i++, arrayIndex++ ) { resultArray1[arrayIndex] = … Read more
[ad_1] You can flatten the internal dict and update it back to the original dict. Eg., product = d.pop(“product”)[0] flattened = {f”product.{key}”: value for key, value in product.items()} d.update(flattened) This should do the trick. 1 [ad_2] solved Flat a list of containing only one dict
[ad_1] When number of other values is just one: $array = [ [‘xx’, 123], [‘xx’, 523], [‘xx’, 783], [‘yy’, 858], [‘yy’, 523], [‘xx’, 235], ]; $result = []; foreach ($array as $row) { list($key, $value) = $row; if (!array_key_exists($key, $result)) { $result[$key] = [$key]; } $result[$key][] = $value; } More generic solution for any number … Read more
[ad_1] This is a good use case for reduce. const rawData = [ { date: ‘3/10/2019’, a: ‘123’, }, { date: ‘3/10/2019’, b: ‘456’, }, { date: ‘3/11/2019’, a: ‘789’, }, { date: ‘3/11/2019’, b: ‘012’, }, { date: ‘3/11/2019’, c: ‘345’, } ]; const groupByDate = array => array.reduce((results, item) => { const current … Read more
[ad_1] This really is a simple one liner via map and delete or map and ES6 destructuring as shown in the 2 examples bellow: var data = [{ “first”: “Linda”, “last”: “Donson”, “salary”: “4000USD” }, { “first”: “Mark”, “last”: “Sullivan”, “salary”: “3500USD” } ] console.log(data.map(x => delete(x.salary) && x)) Also if you are concerned about … Read more