Query to check if a range is made up of consecutive elements
Given an array of n non-consecutive integers and Q queries, the task is to check whether for the given range l and r, the elements are consecutive or not.
Examples:
Input: arr = { 2, 4, 3, 7, 6, 1}, Q = { (1, 3), (3, 5), (5, 6) }
Output: Yes, No, No
Explanation: Array elements from (1, 3) = {2, 4, 3}, (3, 5) = {3, 7, 6}, (5, 6) = {6, 1}
So the answer is Yes, No, No
Input: arr = {1, 2, 3, 4, 5}, Q = { (1, 2), (4, 5), (1, 5) }
Output:Yes, Yes, Yes
Naive Approach:
The sub array between range l and r can be sorted and checked if it contains consecutive elements or not.
Time Complexity: O(q*n*log n)
where n is the size of the array and q is the number of queries.
Efficient approach:
- A segment tree can be used to get the minimum and maximum in a range.
- If the minimum element of a sub array is x and the maximum element is y then we can there is no element less than x and more than y, so we can say that x<=arr[i]<=y.
- As there are no repeated elements, so it can be concluded that if the difference of maximum and minimum element + 1 is equal to the length of the range (r-l+1) then the range consists of consecutive elements.
Below is the implementation of the above approach.
CPP
// C++ implementation of the above approach. #include <bits/stdc++.h> using namespace std; // Segment tree int tree_min[1000001], tree_max[1000001]; // Functions to return // minimum and maximum int min( int a, int b) { return a < b ? a : b; } int max( int a, int b) { return a > b ? a : b; } // Segment tree for minimum element void build_min( int array[], int node, int left, int right) { // If left is equal to right if (left == right) tree_min[node] = array[left]; else { // Divide the segment into equal halves build_min(array, 2 * node, left, (left + right) / 2); build_min(array, 2 * node + 1, (left + right) / 2 + 1, right); // Update the element as minimum // of the divided two subarrays tree_min[node] = min(tree_min[2 * node], tree_min[2 * node + 1]); } } // Query to find a minimum // in a given ranges int query_min( int node, int c_l, int c_r, int l, int r) { // Out of range if (c_l > r || c_r < l) return 1e9; // Within the range completely if (c_l >= l && c_r <= r) return tree_min[node]; else // Divide the range into two halves // Query for each half // and return the minimum of two return min(query_min(2 * node, c_l, (c_l + c_r) / 2, l, r), query_min(2 * node + 1, (c_l + c_r) / 2 + 1, c_r, l, r)); } // Segment tree for maximum element void build_max( int array[], int node, int left, int right) { // If left is equal to right if (left == right) tree_max[node] = array[left]; else { // Divide the segment into equal halves build_max(array, 2 * node, left, (left + right) / 2); build_max(array, 2 * node + 1, (left + right) / 2 + 1, right); // Update the element as maximum // of the divided two subarrays tree_max[node] = max(tree_max[2 * node], tree_max[2 * node + 1]); } } // Query to find maximum // in a given ranges int query_max( int node, int c_l, int c_r, int l, int r) { // Out of range if (c_l > r || c_r < l) return -1; // Within the range completely if (c_l >= l && c_r <= r) return tree_max[node]; else // Divide the range into two halves // and query for each half // and return the maximum of two return max(query_max(2 * node, c_l, (c_l + c_r) / 2, l, r), query_max(2 * node + 1, (c_l + c_r) / 2 + 1, c_r, l, r)); } // Build the tree void init( int array[], int n) { build_min(array, 1, 0, n - 1); build_max(array, 1, 0, n - 1); } // Check if the given range is Consecutive bool isConsecutive( int x, int y, int n) { return ((query_max(1, 0, n - 1, x, y) - query_min(1, 0, n - 1, x, y)) == (y - x)); } // Driver code int main() { int arr[] = { 2, 4, 3, 7, 6, 1 }; int query[][2] = { { 1, 3 }, { 3, 5 }, { 5, 6 } }; int n = sizeof (arr) / sizeof ( int ), q = 3; init(arr, n); for ( int i = 0; i < q; i++) { int l, r; l = query[i][0]; r = query[i][1]; cout << (isConsecutive(l - 1, r - 1, n) ? "Yes" : "No" ) << "\n" ; } return 0; } |
Java
// Java implementation of the above approach. class GFG { // Segment tree static int [] tree_min = new int [ 1000001 ]; static int [] tree_max = new int [ 1000001 ]; // Functions to return // minimum and maximum static int min( int a, int b) { return a < b ? a : b; } static int max( int a, int b) { return a > b ? a : b; } // Segment tree for minimum element static void build_min( int array[], int node, int left, int right) { // If left is equal to right if (left == right) tree_min[node] = array[left]; else { // Divide the segment into equal halves build_min(array, 2 * node, left, (left + right) / 2 ); build_min(array, 2 * node + 1 , (left + right) / 2 + 1 , right); // Update the element as minimum // of the divided two subarrays tree_min[node] = Math.min(tree_min[ 2 * node], tree_min[ 2 * node + 1 ]); } } // Query to find a minimum // in a given ranges static int query_min( int node, int c_l, int c_r, int l, int r) { // Out of range if (c_l > r || c_r < l) return ( int ) 1e9; // Within the range completely if (c_l >= l && c_r <= r) return tree_min[node]; else // Divide the range into two halves // Query for each half // and return the minimum of two return min(query_min( 2 * node, c_l, (c_l + c_r) / 2 , l, r), query_min( 2 * node + 1 , (c_l + c_r) / 2 + 1 , c_r, l, r)); } // Segment tree for maximum element static void build_max( int array[], int node, int left, int right) { // If left is equal to right if (left == right) tree_max[node] = array[left]; else { // Divide the segment into equal halves build_max(array, 2 * node, left, (left + right) / 2 ); build_max(array, 2 * node + 1 , (left + right) / 2 + 1 , right); // Update the element as maximum // of the divided two subarrays tree_max[node] = Math.max(tree_max[ 2 * node], tree_max[ 2 * node + 1 ]); } } // Query to find maximum // in a given ranges static int query_max( int node, int c_l, int c_r, int l, int r) { // Out of range if (c_l > r || c_r < l) return - 1 ; // Within the range completely if (c_l >= l && c_r <= r) return tree_max[node]; else // Divide the range into two halves // and query for each half // and return the maximum of two return Math.max(query_max( 2 * node, c_l, (c_l + c_r) / 2 , l, r), query_max( 2 * node + 1 , (c_l + c_r) / 2 + 1 , c_r, l, r)); } // Build the tree static void init( int array[], int n) { build_min(array, 1 , 0 , n - 1 ); build_max(array, 1 , 0 , n - 1 ); } // Check if the given range is Consecutive static boolean isConsecutive( int x, int y, int n) { return ((query_max( 1 , 0 , n - 1 , x, y) - query_min( 1 , 0 , n - 1 , x, y)) == (y - x)); } // Driver code public static void main(String[] args) { int arr[] = { 2 , 4 , 3 , 7 , 6 , 1 }; int query[][] = { { 1 , 3 }, { 3 , 5 }, { 5 , 6 } }; int n = arr.length, q = 3 ; init(arr, n); for ( int i = 0 ; i < q; i++) { int l, r; l = query[i][ 0 ]; r = query[i][ 1 ]; System.out.print((isConsecutive(l - 1 , r - 1 , n) ? "Yes" : "No" ) + "\n" ); } } } // This code is contributed by PrinciRaj1992 |
Python
# Python3 implementation of the above approach. # Segment tree tree_min = [ 0 ] * 1000001 tree_max = [ 0 ] * 1000001 # Segment tree for minimum element def build_min(array, node,left, right): # If left is equal to right if (left = = right): tree_min[node] = array[left] else : # Divide the segment into equal halves build_min(array, 2 * node, left,(left + right) / / 2 ) build_min(array, 2 * node + 1 ,(left + right) / / 2 + 1 , right) # Update the element as minimum # of the divided two subarrays tree_min[node] = min (tree_min[ 2 * node], tree_min[ 2 * node + 1 ]) # Query to find a minimum # in a given ranges def query_min(node, c_l, c_r, l, r): # Out of range if (c_l > r or c_r < l): return 10 * * 9 # Within the range completely if (c_l > = l and c_r < = r): return tree_min[node] else : # Divide the range into two halves # Query for each half # and return the minimum of two return min (query_min( 2 * node, c_l,(c_l + c_r) / / 2 , l, r), query_min( 2 * node + 1 , (c_l + c_r) / / 2 + 1 ,c_r, l, r)) # Segment tree for maximum element def build_max(array, node, left, right): # If left is equal to right if (left = = right): tree_max[node] = array[left] else : # Divide the segment into equal halves build_max(array, 2 * node, left,(left + right) / / 2 ) build_max(array, 2 * node + 1 ,(left + right) / / 2 + 1 , right) # Update the element as maximum # of the divided two subarrays tree_max[node] = max (tree_max[ 2 * node],tree_max[ 2 * node + 1 ]) # Query to find maximum # in a given ranges def query_max(node, c_l, c_r, l, r): # Out of range if (c_l > r or c_r < l): return - 1 # Within the range completely if (c_l > = l and c_r < = r): return tree_max[node] else : # Divide the range into two halves # and query for each half # and return the maximum of two return max (query_max( 2 * node, c_l,(c_l + c_r) / / 2 , l, r),query_max( 2 * node + 1 ,(c_l + c_r) / / 2 + 1 ,c_r, l, r)) # Build the tree def init(array, n): build_min(array, 1 , 0 , n - 1 ) build_max(array, 1 , 0 , n - 1 ) # Check if the given range is Consecutive def isConsecutive(x, y, n): return ((query_max( 1 , 0 , n - 1 , x, y) - query_min( 1 , 0 , n - 1 , x, y)) = = (y - x)) # Driver code arr = [ 2 , 4 , 3 , 7 , 6 , 1 ] query = [ [ 1 , 3 ] , [ 3 , 5 ], [ 5 , 6 ] ] n = len (arr) q = 3 init(arr, n) for i in range (q): l = query[i][ 0 ] r = query[i][ 1 ] if (isConsecutive(l - 1 , r - 1 , n) ): print ( "Yes" ) else : print ( "No" ) # This code is contributed by mohit kumar 29 |
C#
// C# implementation of the above approach. using System; class GFG { // Segment tree static int [] tree_min = new int [1000001]; static int [] tree_max = new int [1000001]; // Functions to return // minimum and maximum static int min( int a, int b) { return a < b ? a : b; } static int max( int a, int b) { return a > b ? a : b; } // Segment tree for minimum element static void build_min( int []array, int node, int left, int right) { // If left is equal to right if (left == right) tree_min[node] = array[left]; else { // Divide the segment into equal halves build_min(array, 2 * node, left, (left + right) / 2); build_min(array, 2 * node + 1, (left + right) / 2 + 1, right); // Update the element as minimum // of the divided two subarrays tree_min[node] = Math.Min(tree_min[2 * node], tree_min[2 * node + 1]); } } // Query to find a minimum // in a given ranges static int query_min( int node, int c_l, int c_r, int l, int r) { // Out of range if (c_l > r || c_r < l) return ( int ) 1e9; // Within the range completely if (c_l >= l && c_r <= r) return tree_min[node]; else // Divide the range into two halves // Query for each half // and return the minimum of two return min(query_min(2 * node, c_l, (c_l + c_r) / 2, l, r), query_min(2 * node + 1, (c_l + c_r) / 2 + 1, c_r, l, r)); } // Segment tree for maximum element static void build_max( int []array, int node, int left, int right) { // If left is equal to right if (left == right) tree_max[node] = array[left]; else { // Divide the segment into equal halves build_max(array, 2 * node, left, (left + right) / 2); build_max(array, 2 * node + 1, (left + right) / 2 + 1, right); // Update the element as maximum // of the divided two subarrays tree_max[node] = Math.Max(tree_max[2 * node], tree_max[2 * node + 1]); } } // Query to find maximum // in a given ranges static int query_max( int node, int c_l, int c_r, int l, int r) { // Out of range if (c_l > r || c_r < l) return -1; // Within the range completely if (c_l >= l && c_r <= r) return tree_max[node]; else // Divide the range into two halves // and query for each half // and return the maximum of two return Math.Max(query_max(2 * node, c_l, (c_l + c_r) / 2, l, r), query_max(2 * node + 1, (c_l + c_r) / 2 + 1, c_r, l, r)); } // Build the tree static void init( int []array, int n) { build_min(array, 1, 0, n - 1); build_max(array, 1, 0, n - 1); } // Check if the given range is Consecutive static bool isConsecutive( int x, int y, int n) { return ((query_max(1, 0, n - 1, x, y) - query_min(1, 0, n - 1, x, y)) == (y - x)); } // Driver code public static void Main(String[] args) { int []arr = {2, 4, 3, 7, 6, 1}; int [,]query = {{ 1, 3 }, { 3, 5 }, { 5, 6 }}; int n = arr.Length, q = 3; init(arr, n); for ( int i = 0; i < q; i++) { int l, r; l = query[i,0]; r = query[i,1]; Console.Write((isConsecutive(l - 1, r - 1, n) ? "Yes" : "No" ) + "\n" ); } } } // This code is contributed by Rajput-Ji |
Javascript
<script> // JavaScript implementation of the above approach. // Segment tree let tree_min = new Array(1000001); let tree_max = new Array(1000001); // Functions to return // minimum and maximum function min(a,b) { return a < b ? a : b; } function max(a,b) { return a > b ? a : b; } // Segment tree for minimum element function build_min(array,node,left,right) { // If left is equal to right if (left == right) tree_min[node] = array[left]; else { // Divide the segment into equal halves build_min(array, 2 * node, left, Math.floor((left + right) / 2)); build_min(array, 2 * node + 1, Math.floor((left + right) / 2) + 1, right); // Update the element as minimum // of the divided two subarrays tree_min[node] = Math.min(tree_min[2 * node], tree_min[2 * node + 1]); } } // Query to find a minimum // in a given ranges function query_min(node,c_l,c_r,l,r) { // Out of range if (c_l > r || c_r < l) return 1e9; // Within the range completely if (c_l >= l && c_r <= r) return tree_min[node]; else // Divide the range into two halves // Query for each half // and return the minimum of two return min(query_min(2 * node, c_l, Math.floor((c_l + c_r) / 2), l, r), query_min(2 * node + 1, Math.floor((c_l + c_r) / 2) + 1, c_r, l, r)); } // Segment tree for maximum element function build_max(array,node,left,right) { // If left is equal to right if (left == right) tree_max[node] = array[left]; else { // Divide the segment into equal halves build_max(array, 2 * node, left, Math.floor((left + right) / 2)); build_max(array, 2 * node + 1, Math.floor((left + right) / 2) + 1, right); // Update the element as maximum // of the divided two subarrays tree_max[node] = Math.max(tree_max[2 * node], tree_max[2 * node + 1]); } } // Query to find maximum // in a given ranges function query_max(node,c_l,c_r,l,r) { // Out of range if (c_l > r || c_r < l) return -1; // Within the range completely if (c_l >= l && c_r <= r) return tree_max[node]; else // Divide the range into two halves // and query for each half // and return the maximum of two return Math.max(query_max(2 * node, c_l, Math.floor((c_l + c_r) / 2), l, r), query_max(2 * node + 1, Math.floor((c_l + c_r) / 2) + 1, c_r, l, r)); } // Build the tree function init(array,n) { build_min(array, 1, 0, n - 1); build_max(array, 1, 0, n - 1); } // Check if the given range is Consecutive function isConsecutive(x,y,n) { return ((query_max(1, 0, n - 1, x, y) - query_min(1, 0, n - 1, x, y)) == (y - x)); } // Driver code let arr=[ 2, 4, 3, 7, 6, 1]; let query=[[ 1, 3 ], [ 3, 5 ], [ 5, 6 ] ]; let n = arr.length, q = 3; init(arr, n); for (let i = 0; i < q; i++) { let l, r; l = query[i][0]; r = query[i][1]; document.write((isConsecutive(l - 1, r - 1, n) ? "Yes" : "No" ) + "<br>" ); } // This code is contributed by patel2127 </script> |
Yes No No
Time Complexity: O(Q*log(n))