Given an array A[] consisting 0s, 1s and 2s, write a function that sorts A[]. The functions should put all 0s first, then all 1s and all 2s in last.
This is an variation of famous Dutch national flag problem.
The problem was posed with three colours, here `0′, `1′ and `2′. The array is divided into four sections:
- a[1..Lo-1] zeroes (red)
- a[Lo..Mid-] ones (white)
- a[Mid..Hi] unknown
- a[Hi+1..N] twos (blue)
The unknown region is shrunk while maintaining these conditions
- Lo := 1; Mid := 1; Hi := N;
- while Mid <= Hi do
- Invariant: a[1..Lo-1]=0 and a[Lo..Mid-1]=1 and a[Hi+1..N]=2; a[Mid..Hi] are unknown.
- case a[Mid] in
- 0: swap a[Lo] and a[Mid]; Lo++; Mid++
- 1: Mid++
- 2: swap a[Mid] and a[Hi]; Hi–
void sort012(int a[], int arr_size){ int lo = 0; int hi = arr_size - 1; int mid = 0; while(mid <= hi) { switch(a[mid]) { case 0: swap(&a[lo++], &a[mid++]); break; case 1: mid++; break; case 2: swap(&a[mid], &a[hi--]); break; } }}
Given an array and a range [lowVal, highVal], partition the array around the range such that array is divided in three parts.
1) All elements smaller than lowVal come first.
2) All elements in range lowVal to highVVal come next.
3) All elements greater than highVVal appear in the end.
The individual elements of three sets can appear in any order.
1) All elements smaller than lowVal come first.
2) All elements in range lowVal to highVVal come next.
3) All elements greater than highVVal appear in the end.
The individual elements of three sets can appear in any order.
A simple solution is to sort the array. This solution does a lot of extra rearrangements and requires O(n Log n) time.
An efficient solution is based on Dutch National Flag based QuickSort. We traverse given array elements from left. We keep track of two pointers, first (called start in below code) to store next position of smaller element (smaller than range) from beginning; and second (called end in below code) to store next position of greater element from end.
// Partitions arr[0..n-1] around [lowVal..highVal] public static void threeWayPartition(int[] arr, int lowVal, int highVal) { int n = arr.length; // Initialize ext available positions for // smaller (than range) and greater lements int start = 0, end = n-1; // Traverse elements from left for(int i = 0; i < end;) { // If current element is smaller than // range, put it on next available smaller // position. if(arr[i] < lowVal) { int temp = arr[start]; arr[start] = arr[i]; arr[i] = temp; start++; i++; } // If current element is greater than // range, put it on next available greater // position. else if(arr[i] > highVal) { int temp = arr[end]; arr[end] = arr[i]; arr[i] = temp; end--; } else i++; } }
In simple QuickSort algorithm, we select an element as pivot, partition the array around pivot and recur for subarrays on left and right of pivot.
Consider an array which has many redundant elements. For example, {1, 4, 2, 4, 2, 4, 1, 2, 4, 1, 2, 2, 2, 2, 4, 1, 4, 4, 4}. If 4 is picked as pivot in Simple QuickSort, we fix only one 4 and recursively process remaining occurrences.
Consider an array which has many redundant elements. For example, {1, 4, 2, 4, 2, 4, 1, 2, 4, 1, 2, 2, 2, 2, 4, 1, 4, 4, 4}. If 4 is picked as pivot in Simple QuickSort, we fix only one 4 and recursively process remaining occurrences.
The idea of 3 way QuickSort is to process all occurrences of pivot and is based on Dutch National Flag algorithm.
In 3 Way QuickSort, an array arr[l..r] is divided in 3 parts: a) arr[l..i] elements less than pivot. b) arr[i+1..j-1] elements equal to pivot. c) arr[j..r] elements greater than pivot.
void partition(int a[], int l, int r, int &i, int &j){ i = l-1, j = r; int p = l-1, q = r; int v = a[r]; while (true) { // From left, find the first element greater than // or equal to v. This loop will definitely terminate // as v is last element while (a[++i] < v); // From right, find the first element smaller than or // equal to v while (v < a[--j]) if (j == l) break; // If i and j cross, then we are done if (i >= j) break; // Swap, so that smaller goes on left greater goes on right swap(a[i], a[j]); // Move all same left occurrence of pivot to beginning of // array and keep count using p if (a[i] == v) { p++; swap(a[p], a[i]); } // Move all same right occurrence of pivot to end of array // and keep count using q if (a[j] == v) { q--; swap(a[j], a[q]); } } // Move pivot element to its correct index swap(a[i], a[r]); // Move all left same occurrences from beginning // to adjacent to arr[i] j = i-1; for (int k = l; k < p; k++, j--) swap(a[k], a[j]); // Move all right same occurrences from end // to adjacent to arr[i] i = i+1; for (int k = r-1; k > q; k--, i++) swap(a[i], a[k]);}// 3-way partition based quick sortvoid quicksort(int a[], int l, int r){ if (r <= l) return; int i, j; // Note that i and j are passed as reference partition(a, l, r, i, j); // Recur quicksort(a, l, j); quicksort(a, i, r);}
Another Implementation using Dutch National Flag Algorithm
void partition(int a[], int low, int high, int &i, int &j){ // To handle 2 elements if (high - low <= 1) { if (a[high] < a[low]) swap(&a[high], &a[low]); i = low; j = high; return; } int mid = low; int pivot = a[high]; while (mid <= high) { if (a[mid]<pivot) swap(&a[low++], &a[mid++]); else if (a[mid]==pivot) mid++; else if (a[mid]>pivot) swap(&a[mid], &a[high--]); } //update i and j i = low-1; j = mid; //or high-1}// 3-way partition based quick sortvoid quicksort(int a[], int low, int high){ if (low>=high) //1 or 0 elements return; int i, j; // Note that i and j are passed as reference partition(a, low, high, i, j); // Recur two halves quicksort(a, low, i); quicksort(a, j, high);}