If you can assume that all values are different, for each value in tab, its position in the sorted array is the number of values that are strictly lower in tab: Here is how it works: #include <stdio.h> void getOrder(const int tab[], int n, int pos[]) { for (int i = 0; i < n; … Read more
Try this: db.collection.aggregate([ {$unwind: “$carrier”}, {$sort: {“carrier.name”: 1}}, {$group: {_id: “$_id”, carrier: {$push: “$carrier”}}} ]) Mongo Playground solved How to sort 1000000 elements array in alphanumeric in mongodb
First, I think you want your input loop to look something like this – for (int x = 0; x < 10; ++x) { System.out.println(“Enter number: ” + (x + 1)); int num = scan.nextInt(); // <– get user input to // test. // Test for negative first. if (num < 0) { negativeList[neg++] = … Read more
You can easily order by station without understanding the SQL query statements using SQLiteDatabase.query() method Cursor cursor = db.query(“TABLE_NAME”,new String[]{“COLUMN1″,”COLUMN2″,”COLUMN3″},null,null,null,null,”COLUMN1 ASC”); 0 solved Listview Order By Name
ACCEPT ELEMENTS IN A LINKED LIST IN ASCENDING ORDER,but the display function prints the smallest number instead of the whole linked list solved ACCEPT ELEMENTS IN A LINKED LIST IN ASCENDING ORDER,but the display function prints the smallest number instead of the whole linked list
try dic.Values.OrderBy(i=>i.CreatedDateTime) 0 solved How to sort dictionary by DateTime attribute of the value
Below is the sample code to parse your json string // I’ve escaped the double quotes so that it can run, you don’t need to do it because you already have the string let responseStr = “[{\”Loc_District\”:8119,\”districtname\”:\”अजमेर \”,\”District_NameEng\”:\”AJMER\”},{\”Loc_District\”:8104,\”districtname\”:\”अलवर \”,\”District_NameEng\”:\”ALWAR\”},{\”Loc_District\”:8125,\”districtname\”:\”बांसवाड़ा\”,\”District_NameEng\”:\”BANSWARA\”},{\”Loc_District\”:8128,\”districtname\”:\”बारां \”,\”District_NameEng\”:\”BARAN\”},{\”Loc_District\”:8115,\”districtname\”:\”बाड़मेर \”,\”District_NameEng\”:\”BARMER\”},{\”Loc_District\”:8105,\”districtname\”:\”भरतपुर \”,\”District_NameEng\”:\”BHARATPUR\”},{\”Loc_District\”:8122,\”districtname\”:\”भीलवाडा \”,\”District_NameEng\”:\”BHILWARA\”},{\”Loc_District\”:8101,\”districtname\”:\”बीकानेर \”,\”District_NameEng\”:\”BIKANER\”},{\”Loc_District\”:8121,\”districtname\”:\”बून्दी \”,\”District_NameEng\”:\”BUNDI\”},{\”Loc_District\”:8126,\”districtname\”:\”चित्तौड़गढ़ \”,\”District_NameEng\”:\”CHITTORGARH\”},{\”Loc_District\”:8102,\”districtname\”:\”चूरू \”,\”District_NameEng\”:\”CHURU\”},{\”Loc_District\”:8109,\”districtname\”:\”दौसा \”,\”District_NameEng\”:\”DAUSA\”},{\”Loc_District\”:8106,\”districtname\”:\”धौलपुर \”,\”District_NameEng\”:\”DHOLPUR\”},{\”Loc_District\”:8124,\”districtname\”:\”डूंगरपुर \”,\”District_NameEng\”:\”DUNGARPUR\”},{\”Loc_District\”:8099,\”districtname\”:\”गंगानगर \”,\”District_NameEng\”:\”GANGANAGAR\”},{\”Loc_District\”:8100,\”districtname\”:\”हनुमानगढ \”,\”District_NameEng\”:\”HANUMANGARH\”},{\”Loc_District\”:8110,\”districtname\”:\”जयपुर \”,\”District_NameEng\”:\”JAIPUR\”},{\”Loc_District\”:8114,\”districtname\”:\”जैसलमेर \”,\”District_NameEng\”:\”JAISALMER\”},{\”Loc_District\”:8116,\”districtname\”:\”जालोर \”,\”District_NameEng\”:\”JALORE\”},{\”Loc_District\”:8129,\”districtname\”:\”झालावाड … Read more
I think, this is not a sorting but a generation of a new array that accomplish your output, in that case, you could do something like this: 1) First filter all A and all B on different arrays. 2) Then create a new array where progressively you put two elements (or the quantity available) of … Read more
The simplest solution is to split the string into a collection of string then sort that collection and concatenate the result into a string again: you can use a custom comparator using the size of the string and feed it to std::stable_sort with the collection of strings to sort: // Your original string std::string your_string … Read more
After sorting the array, you can simple iterate through each of the array elements and print out the index for each of them. for(int i = 0; i < arr.length; i++) { System.out.println(arr[i] + ” is at the position ” + (i + 1)); } Note that you need to use i + 1 since … Read more
You can define a natural ordering for your class IceCream by implementing the Comparator interface. public class IceCream implements Comparator{ // … final String name; final Date date; public Icecream(String name, Date date){ this.name = name; this.date = date; } public int compare(Object o1, Object o2) { return ((IceCream)o1).date.compareTo(((IceCream)o2).date); } } 3 solved sorting (constructor) … Read more
You can use a TreeMap to keep the items sorted by key and just increment a counter for each occurrence. import java.util.Map; import java.util.Map.Entry; import java.util.TreeMap; public class Counter { public static void main(String[] args) { int[] arr = { 2, 5, 6, 1, 2, 1, 5, 3, 6, 1, 2, 1, 5 }; Map<Integer, … Read more
The reasons for this question getting downvoted are: There’s no effort shown. Tell us what you’ve tried and what you know. We love code. Please show us your code. Even the non-working one. If one thinks about the problem for a while, unanswered questions begin to pop up. For example: Do all the numbers have … Read more
You can try this: for (int i=6; i>1; i–) { for (int j=0; j<i; j++) cout<<“*”; cout<<endl; } Output: ****** ***** **** *** ** 2 solved Print 6,5,4,3,2 characters separated by a new line in C++?
if unsorted[0] <<== unsorted[1] then numsmall = unsorted[a] ^ (eval):51: syntax error, unexpected kTHEN, expecting kEND That little ^ points to first problem here. <<== is not a legal operator in ruby, hence the syntax error. Perhaps you mean “less than or equal to” which is <=? if unsorted[a] <= unsorted[b] Also, indentation will help … Read more