Given two arrays A1[] and A2[], sort A1 in such a way that the relative order among the elements will be same as those are in A2. For the elements not present in A2, append them at last in sorted order.
Input: A1[] = {2, 1, 2, 5, 7, 1, 9, 3, 6, 8, 8}
A2[] = {2, 1, 8, 3}
Output: A1[] = {2, 2, 1, 1, 8, 8, 3, 5, 6, 7, 9}
The code should handle all cases like number of elements in A2[] may be more or less compared to A1[]. A2[] may have some elements which may not be there in A1[] and vice versa is also possible.
https://shawnlincoding.wordpress.com/2015/04/13/sort-array-given-order/
public void sortOrder(int[] array, int[] order){ if(array == null || order == null){ throw new IllegalArgumentException(); } if(array.length == 0 && order.length == 0){ return; } if(array.length != order.length){ return; } int N = array.length; for(int i = 0; i < N; i++){ while(i != order[i]){ swap(array, i, order[i]); swap(order, i, order[i]); } } } private void swap(int[] num, int i, int j){ int temp = num[i]; num[i] = num[j]; num[j] = temp; }Method 1 (Using Sorting and Binary Search)
Let size of A1[] be m and size of A2[] be n.
1) Create a temporary array temp of size m and copy contents of A1[] to it.
2) Create another array visited[] and initialize all entries in it as false. visited[] is used to mark those elements in temp[] which are copied to A1[].
3) Sort temp[]
4) Initialize the output index ind as 0.
5) Do following for every element of A2[i] in A2[]
…..a) Binary search for all occurrences of A2[i] in temp[], if present then copy all occurrences to A1[ind] and increment ind. Also mark the copied elements visited[]
6) Copy all unvisited elements from temp[] to A1[].
1) Create a temporary array temp of size m and copy contents of A1[] to it.
2) Create another array visited[] and initialize all entries in it as false. visited[] is used to mark those elements in temp[] which are copied to A1[].
3) Sort temp[]
4) Initialize the output index ind as 0.
5) Do following for every element of A2[i] in A2[]
…..a) Binary search for all occurrences of A2[i] in temp[], if present then copy all occurrences to A1[ind] and increment ind. Also mark the copied elements visited[]
6) Copy all unvisited elements from temp[] to A1[].
Time complexity: The steps 1 and 2 require O(m) time. Step 3 requires O(mLogm) time. Step 5 requires O(nLogm) time. Therefore overall time complexity is O(m + nLogm).
/* A Binary Search based function to find index of FIRST occurrence of x in arr[]. If x is not present, then it returns -1 */int first(int arr[], int low, int high, int x, int n){ if (high >= low) { int mid = low + (high-low)/2; /* (low + high)/2; */ if ((mid == 0 || x > arr[mid-1]) && arr[mid] == x) return mid; if (x > arr[mid]) return first(arr, (mid + 1), high, x, n); return first(arr, low, (mid -1), x, n); } return -1;}// Sort A1[0..m-1] according to the order defined by A2[0..n-1].void sortAccording(int A1[], int A2[], int m, int n){ // The temp array is used to store a copy of A1[] and visited[] // is used mark the visited elements in temp[]. int temp[m], visited[m]; for (int i=0; i<m; i++) { temp[i] = A1[i]; visited[i] = 0; } // Sort elements in temp sort(temp, temp + m); int ind = 0; // for index of output which is sorted A1[] // Consider all elements of A2[], find them in temp[] // and copy to A1[] in order. for (int i=0; i<n; i++) { // Find index of the first occurrence of A2[i] in temp int f = first(temp, 0, m-1, A2[i], m); // If not present, no need to proceed if (f == -1) continue; // Copy all occurrences of A2[i] to A1[] for (int j = f; (j<m && temp[j]==A2[i]); j++) { A1[ind++] = temp[j]; visited[j] = 1; } } // Now copy all items of temp[] which are not present in A2[] for (int i=0; i<m; i++) if (visited[i] == 0) A1[ind++] = temp[i];}1. Loop through A1[], store the count of every number in a HashMap (key: number, value: count of number) .
2. Loop through A2[], check if it is present in HashMap, if so, put in output array that many times and remove the number from HashMap.
3. Sort the rest of the numbers present in HashMap and put in output array.
http://www.techiedelight.com/custom-sort-sort-elements-array-order-elements-defined-second-array/
The idea is to count frequency of each element of first array and store it in a map. Now for each element of second array, we check if the element is present in the map or not. If it is present in the map, then we print the element n number of times where n is the frequency of that element in first array. We also remove that element from the map so that we are only left with elements that are only present in the first array (but not present in the second array). In order to append them in the end, they need to be sorted.
Note that if we use std::map, we have O(log n) insertion, retrieval time but keys are already ordered, so no need to sort. If we use std::unordered_map, we have O(1) insertion, retrieval time but its keys are unordered, so we need to sort the keys..
void customSort(int arr1[], int arr2[], int m, int n){
// map to store frequency of each element of first array
unordered_map<int, int> freq;
// freq frequency of each element of first array and
// store it in a map.
for (int i = 0; i < m; i++)
freq[arr1[i]]++;
// Note that once we have the frequencies of all elements of
// first array, we can overwrite elements of first array
int index = 0;
// do for every element of second array
for (int i = 0; i < n; i++)
{
// If current element is present in the map, print it n times
// where n is the frequency of that element in first array
while (freq[arr2[i]])
{
arr1[index++] = arr2[i];
freq[arr2[i]]--;
}
// erase the element from map
freq.erase(arr2[i]);
}
// Now we are only left with elements that are only present
// in the first array but not present in the second array
// We need to sort the remaining elements present in the map
// Note - If use std::map, keys already sorted
int i = index;
for (auto it = freq.begin(); it != freq.end(); ++it)
while (it->second--)
arr1[index++] = (it->first);
// sort the remaining elements
sort(arr1 + i, arr1 + index);
}
.Method 2 (Using Self-Balancing Binary Search Tree)
We can also use a self balancing BST like AVL Tree, Red Black Tree, etc. Following are detailed steps.
1) Create a self balancing BST of all elements in A1[]. In every node of BST, also keep track of count of occurrences of the key and a bool field visited which is initialized as false for all nodes.
2) Initialize the output index ind as 0.
3) Do following for every element of A2[i] in A2[]
…..a) Search for A2[i] in the BST, if present then copy all occurrences to A1[ind] and increment ind. Also mark the copied elements visited in the BST node.
4) Do an inorder traversal of BST and copy all unvisited keys to A1[].
http://algorithms.tutorialhorizon.com/sort-an-given-array-in-the-order-defined-by-another-array/
public int [] SortAndSearch(int [] arrA, int [] arrB){
//create an Aux array and copy all the elements of arrA
// create another boolean array, visited[] of size arrA[]
// Initialize visited[] = false
// Sort the Aux array using Merge sort.
// Navigate through the arrB, taking one element at a time, say x
// 1. perform binary search of x on Aux array and find the first occurrence of x
// 2. if x is found, copy it to arrA, make visited[] = true
// 3. Do linear navigation on Aux array till you find x, copy all these to arrA and mark visited[] = true
// When arrB is over, copy rest of the elements from Aux to arrA.
int oIndex = -1;
boolean [] visited = new boolean [arrA.length];
for(int i=0;i<visited.length;i++){
visited[i] = false;
}
int [] Aux = new int[arrA.length];
for(int i=0;i<arrA.length;i++){
Aux[i] = arrA[i];
}
System.out.println("Original Array : ");
displayArray(arrA);
System.out.println("\nDefined Array : ");
displayArray(arrB);
MergeSort m = new MergeSort(Aux);
Aux = m.mergeSorting();
for(int i=0;i<arrB.length;i++){
int x = arrB[i];
int index = findFirstOccurrence(Aux, x, 0, Aux.length-1);
if(index>=0){
arrA[++oIndex]=Aux[index];
visited[index] = true;
//oIndex++;
while(Aux[++index]==x){
arrA[++oIndex]=Aux[index];
visited[index] = true;
}
}
}
for(int i=0;i<Aux.length;i++){
if(visited[i]==false){
arrA[++oIndex] = Aux[i];
}
}
return arrA;
}
public int findFirstOccurrence(int [] arrA, int x, int start, int end){
if(end>=start){
int mid = (start+end)/2;
if((mid==0||(arrA[mid-1]<x)) && arrA[mid]==x){
return mid;
}else if(arrA[mid]<x){
return findFirstOccurrence(arrA, x, mid+1, end);
}else{
return findFirstOccurrence(arrA, x, start, mid-1);
}
}else return -1;
}
public int [] usingHashing(int [] arrA, int [] arrB){
//Insert all the elements of arrA in hash Table,
//Element as key and its count as its value
//Navigate through arrB, check element's presence in Hash table
//if it is present then takes its value (count) and insert into arrA.
//Once arrB is completed, take rest of the elements from Hash table
// Sort Them and insert into arrB.
int resIndex = -1;
Hashtable<Integer, Integer> h = new Hashtable<>();
for(int i=0;i<arrA.length;i++){
if(h.containsKey(arrA[i])){
int count = h.get(arrA[i]);
count++;
h.put(arrA[i], count);
}else{
h.put(arrA[i], 1);
}
}
for(int i=0;i<arrB.length;i++){
if(h.containsKey(arrB[i])){
int count = h.get(arrB[i]);
while(count>0){
arrA[++resIndex] = arrB[i];
count--;
}
h.remove(arrB[i]);
}
}
ArrayList<Integer> al = new ArrayList<Integer>();
Set<Integer> keys = h.keySet();
for(Integer x:keys){
al.add(x);
}
Collections.sort(al);
Iterator<Integer> it = al.iterator();
while(it.hasNext()){
arrA[++resIndex] = it.next();
}
return arrA;
}
Method 4 (By Writing a Customized Compare Method)
http://www.techiedelight.com/custom-sort-sort-elements-array-order-elements-defined-second-array/
We can also write a custom compare method to solve this problem. Let the two element to be compared are x and y. Then
- If both x and y are present in the second array, then the element with lower index in the second array should come first.
- If only one of x or y is present in the second array, then the element which is present in the second array should come first.
- If both elements are not present in second array, then the default ordering will be considered.
{
Map<Integer, Integer> map;
public CustomComparator(Map<Integer, Integer> map)
{
this.map = map;
}
public int compare(Integer x, Integer y)
{
if (map.containsKey(x) && map.containsKey(y))
return map.get(x) - map.get(y);
else if (map.containsKey(y))
return 1;
else if (map.containsKey(x))
return -1;
else
return x - y;
}
}
// Wrapper class used
Integer[] arr1 = { 5, 8, 9, 3, 5, 7, 1, 3, 4, 9, 3, 5, 1, 8, 4 };
Integer[] arr2 = { 3, 5, 7, 2 };
Map<Integer, Integer> map = new HashMap<Integer, Integer>();
for (int i = 0; i < arr2.length; i++)
map.put(arr2[i], i);
// Arrays.sort method doesn't work with primitive array
// when used with user defined comparator
Arrays.sort(arr1, new CustomComparator(map));
int search(int key){ int i=0, idx = 0; for (i=0; i<size; i++) if (A2[i] == key) return i; return -1;}// A custom comapre method to compare elements of A1[] according// to the order defined by A2[].int compareByA2(const void * a, const void * b){ int idx1 = search(*(int*)a); int idx2 = search(*(int*)b); if (idx1 != -1 && idx2 != -1) return idx1 - idx2; else if(idx1 != -1) return -1; else if(idx2 != -1) return 1; else return ( *(int*)a - *(int*)b );}// This method mainly uses qsort to sort A1[] according to A2[]void sortA1ByA2(int A1[], int size1){ qsort(A1, size1, sizeof (int), compareByA2);}