Find a partition point in array - GeeksforGeeks
Given an unsorted array of integers. Find an element such that all the elements to its left are smaller and to its right are greater. Print -1 if no such element exists.
Note that there can be more than one such elements. For example an array which is sorted in increasing order all elements follow the property. We need to find only one such element.
Read full article from Find a partition point in array - GeeksforGeeks
Given an unsorted array of integers. Find an element such that all the elements to its left are smaller and to its right are greater. Print -1 if no such element exists.
Note that there can be more than one such elements. For example an array which is sorted in increasing order all elements follow the property. We need to find only one such element.
Efficient solution take O(n) time.
- Create an auxiliary array ‘GE[]’. GE[] should store the element which is greater than A[i] and is on left side of A[i].
- Create an another Auxliary array ‘SE[]’. SE[i] should store the element which is smaller than A[i] and is on right side of A[i].
- Find element in array that hold condition GE[i-1] < A[i] < SE[i+1].
int FindElement(int A[], int n){ // Create an array 'SE[]' that will store smaller // element on right side. int SE[n]; // Create an another array 'GE[]' that will store // greatest element on left side. int GE[n]; // initalize first and last index of SE[] , GE[] GE[0] = A[0]; SE[n-1] = A[n-1]; // store greatest element from left to right for (int i=1; i<n ; i++) { if (GE[i-1] < A[i]) GE[i] = A[i]; else GE[i] = GE[i-1]; } // store smallest element from right to left for (int i = n-2 ; i >= 0 ; i-- ) { if (A[i] < SE[i+1]) SE[i] = A[i]; else SE[i] = SE[i+1]; } // Now find a number which is greater then all // elements at it's left and smaller the all // then elements to it's right for (int j =0 ; j <n ; j++) { if ( ( j == 0 && A[j] < SE[j+1] ) || ( j == n-1 && A[j] > GE[j-1] ) || ( A[j] < SE[j+1] && A[j] > GE[j-1] ) ) return A[j]; } return -1;}
Simple solution takes O(n2). Idea is to pick each array element one by one and for each element we have to check it is greater than all the elements to its left side and smaller than all the elements to its right side.
int FindElement( int A[] , int n ){ // traverse array elements for (int i=0; i < n ; i++) { // If we found that such number int flag = 0 ; // check All the elements on its left are smaller for (int j = 0 ; j < i ; j++ ) if (A[j] >= A[i] ) { flag = 1 ; break; } // check All the elements on its right are Greater for( int j = i + 1 ; j < n; j++ ) if( A[j] <= A[i] ) { flag = 1 ; break; } // If flag == 0 indicates we found that number if (flag == 0) return A[i]; } return -1;}