Find the Minimum length Unsorted Subarray, sorting which makes the complete array sorted | GeeksforGeeks
Given an unsorted array arr[0..n-1] of size n, find the minimum length subarray arr[s..e] such that sorting this subarray makes the whole array sorted.
http://www.android100.org/html/201501/10/97072.html
void findUnsortedSequence(int[] array) {
// find left subsequence
int end left= findEndOfleftSubsequence(array);
if (end_left >= array.length - 1) return; // Already sorted
// find right subsequence
int start_right = findStartOfRightSubsequence(array);
// get min and max
int max_index = end_left; // max of left side
int min_index = start_right; // min of right side
for (int i= end_left + 1; i < start_right; i++) {
if (array[i] < array[min_index]) min index = i;
if (array[i] > array[max_index]) max_index = i;
}
// slide left until less than array[min_index]
int left_index = shrinkLeft(array, min_index, end_left);
// slide right until greater than array[max_index]
int right_index = shrinkRight(array, max_index, start_right);
System.out.println(left_index +" "+ right_index);
}
int findEndOfLeftSubsequence(int[] array) {
for (int i= 1; i < array.length; i++) { //
if (array[i] < array[i - 1]) return i - 1;
}
return array.length - 1;
}
int findStartOfRightSubsequence(int[] array) {
for (int i= array.length - 2; i >= 0; i--) {
if (array[i] > array[i + 1]) return i + 1;
}
return 0;
}
int shrinkleft(int[] array, int min_index, int start) {
int comp= array[min_index];
for (inti= start - 1; i >= 0; i--) {
if (array[i] <= comp) return i + 1;
}
return 0;
}
int shrinkRight(int[] array, int max_index, int start) {
int comp = array[max_index];
for (int i= start; i < array.length; i++) {
if (array[i] >= comp) return i - 1;
}
return array.length - 1;
}
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http://blueocean-penn.blogspot.com/2015/01/given-unsorted-list-find-minimum-window.html
Read full article from Find the Minimum length Unsorted Subarray, sorting which makes the complete array sorted | GeeksforGeeks
Given an unsorted array arr[0..n-1] of size n, find the minimum length subarray arr[s..e] such that sorting this subarray makes the whole array sorted.
1) If the input array is [10, 12, 20, 30, 25, 40, 32, 31, 35, 50, 60], your program should be able to find that the subarray lies between the indexes 3 and 8.
1) Find the candidate unsorted subarray
a) Scan from left to right and find the first element which is greater than the next element. Let s be the index of such an element. In the above example 1, s is 3 (index of 30).
b) Scan from right to left and find the first element (first in right to left order) which is smaller than the next element (next in right to left order). Let e be the index of such an element. In the above example 1, e is 7 (index of 31).
a) Scan from left to right and find the first element which is greater than the next element. Let s be the index of such an element. In the above example 1, s is 3 (index of 30).
b) Scan from right to left and find the first element (first in right to left order) which is smaller than the next element (next in right to left order). Let e be the index of such an element. In the above example 1, e is 7 (index of 31).
2) Check whether sorting the candidate unsorted subarray makes the complete array sorted or not. If not, then include more elements in the subarray.
a) Find the minimum and maximum values in arr[s..e]. Let minimum and maximum values be min andmax. min and max for [30, 25, 40, 32, 31] are 25 and 40 respectively.
b) Find the first element (if there is any) in arr[0..s-1] which is greater than min, change s to index of this element. There is no such element in above example 1.
c) Find the last element (if there is any) in arr[e+1..n-1] which is smaller than max, change e to index of this element. In the above example 1, e is changed to 8 (index of 35)
a) Find the minimum and maximum values in arr[s..e]. Let minimum and maximum values be min andmax. min and max for [30, 25, 40, 32, 31] are 25 and 40 respectively.
b) Find the first element (if there is any) in arr[0..s-1] which is greater than min, change s to index of this element. There is no such element in above example 1.
c) Find the last element (if there is any) in arr[e+1..n-1] which is smaller than max, change e to index of this element. In the above example 1, e is changed to 8 (index of 35)
3) Print s and e.
找到heading的最长递增序列.
找到tailing的最长的递增序列.
After that:
用中间部分的min去shrink左边.
用中间部分的max去shrink右边.
| public static void findMinimalUnsortedRange(int[] a){ int leftEnd = findLeftEnd(a); | |
| int rightStart = findRightStart(a); | |
| int begin = 0; | |
| int end = 0; | |
| //THIS INITIALIZATION IS CRUCIAL! otherwise, there will be problem considering input: {1,2,4,7,10,11,81, 7, 12,6,7,16,18,19}; | |
| int min = a[rightStart]; | |
| int max = a[leftEnd]; | |
| //find the min and max of the middle part | |
| for(int i = leftEnd+1; i <= rightStart-1; i++){ | |
| min = a[i] < min ? a[i] : min; | |
| max = a[i] > max ? a[i] : max; | |
| } | |
| //shrink the left part if the largest value of left part > min of middle part | |
| while(leftEnd >= 0 && a[leftEnd] > min){ | |
| leftEnd--; | |
| } | |
| begin = leftEnd+1; | |
| //shrink the right part if the min value of right part < max of middle part | |
| while(rightStart <= a.length-1 && a[rightStart] < max){ | |
| rightStart++; | |
| } | |
| end = rightStart-1; | |
| System.out.println("begin is " + begin); | |
| System.out.println("end is " + end); | |
| } | |
| public static int findLeftEnd(int[] a){ | |
| int len = a.length; | |
| for(int i = 0; i < len-1; i++){ | |
| if(a[i] > a[i+1]){ | |
| return i; | |
| } | |
| } | |
| return len-1; | |
| } | |
| public static int findRightStart(int[] a){ | |
| int len = a.length; | |
| for(int i = len-1; i > 0; i--){ | |
| if(a[i] < a[i-1]){ | |
| return i; | |
| } | |
| } | |
| return 0; | |
| } |
void printUnsorted(int arr[], int n){ int s = 0, e = n-1, i, max, min; // step 1(a) of above algo for (s = 0; s < n-1; s++) { if (arr[s] > arr[s+1]) break; } if (s == n-1) { printf("The complete array is sorted"); return; } // step 1(b) of above algo for(e = n - 1; e > 0; e--) { if(arr[e] < arr[e-1]) break; } // step 2(a) of above algo max = arr[s]; min = arr[s]; for(i = s + 1; i <= e; i++) { if(arr[i] > max) max = arr[i]; if(arr[i] < min) min = arr[i]; } // step 2(b) of above algo for( i = 0; i < s; i++) { if(arr[i] > min) { s = i; break; } } // step 2(c) of above algo for( i = n -1; i >= e+1; i--) { if(arr[i] < max) { e = i; break; } } // step 3 of above algo printf(" The unsorted subarray which makes the given array " " sorted lies between the indees %d and %d", s, e); return;}http://www.android100.org/html/201501/10/97072.html
void findUnsortedSequence(int[] array) {
// find left subsequence
int end left= findEndOfleftSubsequence(array);
if (end_left >= array.length - 1) return; // Already sorted
// find right subsequence
int start_right = findStartOfRightSubsequence(array);
// get min and max
int max_index = end_left; // max of left side
int min_index = start_right; // min of right side
for (int i= end_left + 1; i < start_right; i++) {
if (array[i] < array[min_index]) min index = i;
if (array[i] > array[max_index]) max_index = i;
}
// slide left until less than array[min_index]
int left_index = shrinkLeft(array, min_index, end_left);
// slide right until greater than array[max_index]
int right_index = shrinkRight(array, max_index, start_right);
System.out.println(left_index +" "+ right_index);
}
int findEndOfLeftSubsequence(int[] array) {
for (int i= 1; i < array.length; i++) { //
if (array[i] < array[i - 1]) return i - 1;
}
return array.length - 1;
}
int findStartOfRightSubsequence(int[] array) {
for (int i= array.length - 2; i >= 0; i--) {
if (array[i] > array[i + 1]) return i + 1;
}
return 0;
}
int shrinkleft(int[] array, int min_index, int start) {
int comp= array[min_index];
for (inti= start - 1; i >= 0; i--) {
if (array[i] <= comp) return i + 1;
}
return 0;
}
int shrinkRight(int[] array, int max_index, int start) {
int comp = array[max_index];
for (int i= start; i < array.length; i++) {
if (array[i] >= comp) return i - 1;
}
return array.length - 1;
}
http://blueocean-penn.blogspot.com/2015/01/given-unsorted-list-find-minimum-window.html
Read full article from Find the Minimum length Unsorted Subarray, sorting which makes the complete array sorted | GeeksforGeeks
