C++ Program for Check if an array is sorted and rotated
Last Updated :
20 Apr, 2023
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Given an array of N distinct integers. The task is to write a program to check if this array is sorted and rotated counter-clockwise. A sorted array is not considered as sorted and rotated, i.e., there should at least one rotation.
Examples:
Input : arr[] = { 3, 4, 5, 1, 2 }
Output : YES
The above array is sorted and rotated.
Sorted array: {1, 2, 3, 4, 5}.
Rotating this sorted array clockwise
by 3 positions, we get: { 3, 4, 5, 1, 2}
Input: arr[] = {7, 9, 11, 12, 5}
Output: YES
Input: arr[] = {1, 2, 3}
Output: NO
Input: arr[] = {3, 4, 6, 1, 2, 5}
Output: NO
Approach:
- Find the minimum element in the array.
- Now, if the array is sorted and then rotate all the elements before the minimum element will be in increasing order and all elements after the minimum element will also be in increasing order.
- Check if all elements before minimum element are in increasing order.
- Check if all elements after minimum element are in increasing order.
- Check if the last element of the array is smaller than the starting element.
- If all of the above three conditions satisfies then print YES otherwise print NO.
Below is the implementation of the above idea:
// CPP program to check if an array is
// sorted and rotated clockwise
#include <climits>
#include <iostream>
using namespace std;
// Function to check if an array is
// sorted and rotated clockwise
void checkIfSortRotated(int arr[], int n)
{
int minEle = INT_MAX;
int maxEle = INT_MIN;
int minIndex = -1;
// Find the minimum element
// and it's index
for (int i = 0; i < n; i++) {
if (arr[i] < minEle) {
minEle = arr[i];
minIndex = i;
}
}
int flag1 = 1;
// Check if all elements before minIndex
// are in increasing order
for (int i = 1; i < minIndex; i++) {
if (arr[i] < arr[i - 1]) {
flag1 = 0;
break;
}
}
int flag2 = 1;
// Check if all elements after minIndex
// are in increasing order
for (int i = minIndex + 1; i < n; i++) {
if (arr[i] < arr[i - 1]) {
flag2 = 0;
break;
}
}
// Check if last element of the array
// is smaller than the element just
// starting element of the array
// for arrays like [3,4,6,1,2,5] - not circular array
if (flag1 && flag2 && (arr[n - 1] < arr[0]))
cout << "YES";
else
cout << "NO";
}
// Driver code
int main()
{
int arr[] = { 3, 4, 5, 1, 2 };
int n = sizeof(arr) / sizeof(arr[0]);
// Function Call
checkIfSortRotated(arr, n);
return 0;
}
// CPP program to check if an array is// sorted and rotated clockwiseusing namespace std;// Function to check if an array is// sorted and rotated clockwisevoid checkIfSortRotated(int arr[], int n){ int minEle = INT_MAX; int maxEle = INT_MIN; int minIndex = -1; // Find the minimum element // and it's index for (int i = 0; i < n; i++) { if (arr[i] < minEle) { minEle = arr[i]; minIndex = i; } } int flag1 = 1; // Check if all elements before minIndex // are in increasing order for (int i = 1; i < minIndex; i++) { if (arr[i] < arr[i - 1]) { flag1 = 0; break; } } int flag2 = 1; // Check if all elements after minIndex // are in increasing order for (int i = minIndex + 1; i < n; i++) { if (arr[i] < arr[i - 1]) { flag2 = 0; break; } } // Check if last element of the array // is smaller than the element just // starting element of the array // for arrays like [3,4,6,1,2,5] - not circular array if (flag1 && flag2 && (arr[n - 1] < arr[0])) cout << "YES"; else cout << "NO";}// Driver codeint main(){ int arr[] = { 3, 4, 5, 1, 2 }; int n = sizeof(arr) / sizeof(arr[0]); // Function Call checkIfSortRotated(arr, n); return 0;}Output
YES
Time Complexity: O(N), where N represents the size of the given array.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
Approach Name: Linear Search
Steps:
- Traverse the array from the second element to the last element and check if the current element is smaller than the previous element. If so, then the array is rotated.
- If the array is rotated, then find the index of the minimum element in the array. This can be done by traversing the array again and keeping track of the minimum element and its index.
- Check if the array is sorted by comparing the elements starting from the minimum element to the previous element in a circular manner. If all elements are in non-descending order, then the array is sorted and rotated.
#include <iostream>
using namespace std;
bool isSortedAndRotated(int arr[], int n) {
bool rotated = false;
int min_index = 0;
int min_element = arr[0];
for (int i = 1; i < n; i++) {
if (arr[i] < arr[i - 1]) {
rotated = true;
}
if (arr[i] < min_element) {
min_index = i;
min_element = arr[i];
}
}
if (!rotated) {
return false;
}
for (int i = 1; i < n; i++) {
int index = (min_index + i) % n;
int prev_index = (min_index + i - 1) % n;
if (arr[index] < arr[prev_index]) {
return false;
}
}
return true;
}
int main() {
int arr1[] = { 3, 4, 5, 1, 2 };
int arr2[] = { 7, 9, 11, 12, 5 };
int arr3[] = { 1, 2, 3 };
int arr4[] = { 3, 4, 6, 1, 2, 5 };
int n1 = sizeof(arr1) / sizeof(arr1[0]);
int n2 = sizeof(arr2) / sizeof(arr2[0]);
int n3 = sizeof(arr3) / sizeof(arr3[0]);
int n4 = sizeof(arr4) / sizeof(arr4[0]);
if (isSortedAndRotated(arr1, n1)) {
cout << "YES\n";
} else {
cout << "NO\n";
}
if (isSortedAndRotated(arr2, n2)) {
cout << "YES\n";
} else {
cout << "NO\n";
}
if (isSortedAndRotated(arr3, n3)) {
cout << "YES\n";
} else {
cout << "NO\n";
}
if (isSortedAndRotated(arr4, n4)) {
cout << "YES\n";
} else {
cout << "NO\n";
}
return 0;
}
using namespace std;bool isSortedAndRotated(int arr[], int n) { bool rotated = false; int min_index = 0; int min_element = arr[0]; for (int i = 1; i < n; i++) { if (arr[i] < arr[i - 1]) { rotated = true; } if (arr[i] < min_element) { min_index = i; min_element = arr[i]; } } if (!rotated) { return false; } for (int i = 1; i < n; i++) { int index = (min_index + i) % n; int prev_index = (min_index + i - 1) % n; if (arr[index] < arr[prev_index]) { return false; } } return true;}int main() { int arr1[] = { 3, 4, 5, 1, 2 }; int arr2[] = { 7, 9, 11, 12, 5 }; int arr3[] = { 1, 2, 3 }; int arr4[] = { 3, 4, 6, 1, 2, 5 }; int n1 = sizeof(arr1) / sizeof(arr1[0]); int n2 = sizeof(arr2) / sizeof(arr2[0]); int n3 = sizeof(arr3) / sizeof(arr3[0]); int n4 = sizeof(arr4) / sizeof(arr4[0]); if (isSortedAndRotated(arr1, n1)) { cout << "YES\n"; } else { cout << "NO\n"; } if (isSortedAndRotated(arr2, n2)) { cout << "YES\n"; } else { cout << "NO\n"; } if (isSortedAndRotated(arr3, n3)) { cout << "YES\n"; } else { cout << "NO\n"; } if (isSortedAndRotated(arr4, n4)) { cout << "YES\n"; } else { cout << "NO\n"; } return 0;}Output
YES YES NO NO
Time Complexity: O(n), where n is the size of the input array.
Auxiliary Space: O(1)
Please refer complete article on Check if an array is sorted and rotated for more details!
