Find smallest range containing elements from k lists
Given k sorted lists of integers of size n each, find the smallest range that includes at least one element from each of the k lists. If more than one smallest range is found, print any one of them.
Example:
Input: K = 3 arr1[] : [4, 7, 9, 12, 15] arr2[] : [0, 8, 10, 14, 20] arr3[] : [6, 12, 16, 30, 50] Output: The smallest range is [6 8] Explanation: Smallest range is formed by number 7 from the first list, 8 from second list and 6 from the third list. Input: k = 3 arr1[] : [4, 7] arr2[] : [1, 2] arr3[] : [20, 40] Output: The smallest range is [2 20] Explanation:The range [2, 20] contains 2, 4, 7, 20 which contains element from all the three arrays.
Naive Approach: Given K sorted list, find a range where there is at least one element from every list. The idea to solve the problem is very simple, keep k pointers which will constitute the elements in the range, by taking the min and max of the k elements the range can be formed. Initially, all the pointers will point to the start of all the k arrays. Store the range max to min. If the range has to be minimized then either the minimum value has to be increased or the maximum value has to be decreased. To decrease the maximum value we have to move our pointer of current maximum to the left and since we are currently at 0 the index of every list so we can’t move our pointer to left, hence we can’t decrease the current max. So, the only possible option to get a better range is to increase the current minimum. To continue increasing the minimum value, increase the pointer of the list containing the minimum value and update the range until one of the lists exhausts.
- Algorithm:
- Create an extra space ptr of length k to store the pointers and a variable minrange initialized to a maximum value.
- Initially the index of every list is 0, therefore initialize every element of ptr[0..k] to 0, the array ptr will store the index of the elements in the range.
- Repeat the following steps until at least one list exhausts:
- Now find the minimum and maximum value among the current elements of all the list pointed by the ptr[0…k] array.
- Now update the minrange if current (max-min) is less than minrange.
- increment the pointer pointing to current minimum element.
- Implementation:
C++
// C++ program to finds out smallest range that includes // elements from each of the given sorted lists. #include <bits/stdc++.h> using namespace std; #define N 5 // array for storing the current index of list i int ptr[501]; // This function takes an k sorted lists in the form of // 2D array as an argument. It finds out smallest range // that includes elements from each of the k lists. void findSmallestRange( int arr[][N], int n, int k) { int i, minval, maxval, minrange, minel, maxel, flag, minind; // initializing to 0 index; for (i = 0; i <= k; i++) ptr[i] = 0; minrange = INT_MAX; while (1) { // for maintaining the index of list containing the minimum element minind = -1; minval = INT_MAX; maxval = INT_MIN; flag = 0; // iterating over all the list for (i = 0; i < k; i++) { // if every element of list[i] is traversed then break the loop if (ptr[i] == n) { flag = 1; break ; } // find minimum value among all the list elements pointing by the ptr[] array if (ptr[i] < n && arr[i][ptr[i]] < minval) { minind = i; // update the index of the list minval = arr[i][ptr[i]]; } // find maximum value among all the list elements pointing by the ptr[] array if (ptr[i] < n && arr[i][ptr[i]] > maxval) { maxval = arr[i][ptr[i]]; } } // if any list exhaust we will not get any better answer, so break the while loop if (flag) break ; ptr[minind]++; // updating the minrange if ((maxval - minval) < minrange) { minel = minval; maxel = maxval; minrange = maxel - minel; } } printf ( "The smallest range is [%d, %d]\n" , minel, maxel); } // Driver program to test above function int main() { int arr[][N] = { { 4, 7, 9, 12, 15 }, { 0, 8, 10, 14, 20 }, { 6, 12, 16, 30, 50 } }; int k = sizeof (arr) / sizeof (arr[0]); findSmallestRange(arr, N, k); return 0; } // This code is contributed by Aditya Krishna Namdeo |
Java
// Java program to finds out smallest range that includes // elements from each of the given sorted lists. class GFG { static final int N = 5 ; // array for storing the current index of list i static int ptr[] = new int [ 501 ]; // This function takes an k sorted lists in the form of // 2D array as an argument. It finds out smallest range // that includes elements from each of the k lists. static void findSmallestRange( int arr[][], int n, int k) { int i, minval, maxval, minrange, minel = 0 , maxel = 0 , flag, minind; // initializing to 0 index; for (i = 0 ; i <= k; i++) { ptr[i] = 0 ; } minrange = Integer.MAX_VALUE; while ( true ) { // for maintaining the index of list containing the minimum element minind = - 1 ; minval = Integer.MAX_VALUE; maxval = Integer.MIN_VALUE; flag = 0 ; // iterating over all the list for (i = 0 ; i < k; i++) { // if every element of list[i] is traversed then break the loop if (ptr[i] == n) { flag = 1 ; break ; } // find minimum value among all the list elements pointing by the ptr[] array if (ptr[i] < n && arr[i][ptr[i]] < minval) { minind = i; // update the index of the list minval = arr[i][ptr[i]]; } // find maximum value among all the list elements pointing by the ptr[] array if (ptr[i] < n && arr[i][ptr[i]] > maxval) { maxval = arr[i][ptr[i]]; } } // if any list exhaust we will not get any better answer, so break the while loop if (flag == 1 ) { break ; } ptr[minind]++; // updating the minrange if ((maxval - minval) < minrange) { minel = minval; maxel = maxval; minrange = maxel - minel; } } System.out.printf( "The smallest range is [%d, %d]\n" , minel, maxel); } // Driver program to test above function public static void main(String[] args) { int arr[][] = { { 4 , 7 , 9 , 12 , 15 }, { 0 , 8 , 10 , 14 , 20 }, { 6 , 12 , 16 , 30 , 50 } }; int k = arr.length; findSmallestRange(arr, N, k); } } // this code contributed by Rajput-Ji |
Python
# Python3 program to finds out # smallest range that includes # elements from each of the # given sorted lists. N = 5 # array for storing the # current index of list i ptr = [ 0 for i in range ( 501 )] # This function takes an k sorted # lists in the form of 2D array as # an argument. It finds out smallest # range that includes elements from # each of the k lists. def findSmallestRange(arr, n, k): i, minval, maxval, minrange, minel, maxel, flag, minind = 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 # initializing to 0 index for i in range (k + 1 ): ptr[i] = 0 minrange = 10 * * 9 while ( 1 ): # for maintaining the index of list # containing the minimum element minind = - 1 minval = 10 * * 9 maxval = - 10 * * 9 flag = 0 # iterating over all the list for i in range (k): # if every element of list[i] is # traversed then break the loop if (ptr[i] = = n): flag = 1 break # find minimum value among all the list # elements pointing by the ptr[] array if (ptr[i] < n and arr[i][ptr[i]] < minval): minind = i # update the index of the list minval = arr[i][ptr[i]] # find maximum value among all the # list elements pointing by the ptr[] array if (ptr[i] < n and arr[i][ptr[i]] > maxval): maxval = arr[i][ptr[i]] # if any list exhaust we will # not get any better answer, # so break the while loop if (flag): break ptr[minind] + = 1 # updating the minrange if ((maxval - minval) < minrange): minel = minval maxel = maxval minrange = maxel - minel print ( "The smallest range is [" , minel, maxel, "]" ) # Driver code arr = [ [ 4 , 7 , 9 , 12 , 15 ], [ 0 , 8 , 10 , 14 , 20 ], [ 6 , 12 , 16 , 30 , 50 ] ] k = len (arr) findSmallestRange(arr, N, k) # This code is contributed by mohit kumar |
C#
// C# program to finds out smallest // range that includes elements from // each of the given sorted lists. using System; class GFG { static int N = 5; // array for storing the current index of list i static int [] ptr = new int [501]; // This function takes an k sorted // lists in the form of 2D array as // an argument. It finds out smallest range // that includes elements from each of the k lists. static void findSmallestRange( int [, ] arr, int n, int k) { int i, minval, maxval, minrange, minel = 0, maxel = 0, flag, minind; // initializing to 0 index; for (i = 0; i <= k; i++) { ptr[i] = 0; } minrange = int .MaxValue; while ( true ) { // for maintaining the index of // list containing the minimum element minind = -1; minval = int .MaxValue; maxval = int .MinValue; flag = 0; // iterating over all the list for (i = 0; i < k; i++) { // if every element of list[i] // is traversed then break the loop if (ptr[i] == n) { flag = 1; break ; } // find minimum value among all the // list elements pointing by the ptr[] array if (ptr[i] < n && arr[i, ptr[i]] < minval) { minind = i; // update the index of the list minval = arr[i, ptr[i]]; } // find maximum value among all the // list elements pointing by the ptr[] array if (ptr[i] < n && arr[i, ptr[i]] > maxval) { maxval = arr[i, ptr[i]]; } } // if any list exhaust we will // not get any better answer, // so break the while loop if (flag == 1) { break ; } ptr[minind]++; // updating the minrange if ((maxval - minval) < minrange) { minel = minval; maxel = maxval; minrange = maxel - minel; } } Console.WriteLine( "The smallest range is" + "[{0}, {1}]\n" , minel, maxel); } // Driver code public static void Main(String[] args) { int [, ] arr = { { 4, 7, 9, 12, 15 }, { 0, 8, 10, 14, 20 }, { 6, 12, 16, 30, 50 } }; int k = arr.GetLength(0); findSmallestRange(arr, N, k); } } // This code has been contributed by 29AjayKumar |
Javascript
<script> // Javascript program to finds out smallest range that includes // elements from each of the given sorted lists. let N = 5; // array for storing the current index of list i let ptr= new Array(501); // This function takes an k sorted lists in the form of // 2D array as an argument. It finds out smallest range // that includes elements from each of the k lists. function findSmallestRange(arr,n,k) { let i, minval, maxval, minrange, minel = 0, maxel = 0, flag, minind; // initializing to 0 index; for (i = 0; i <= k; i++) { ptr[i] = 0; } minrange = Number.MAX_VALUE; while ( true ) { // for maintaining the index of list containing the minimum element minind = -1; minval = Number.MAX_VALUE; maxval = Number.MIN_VALUE; flag = 0; // iterating over all the list for (i = 0; i < k; i++) { // if every element of list[i] is traversed then break the loop if (ptr[i] == n) { flag = 1; break ; } // find minimum value among all the list elements pointing by the ptr[] array if (ptr[i] < n && arr[i][ptr[i]] < minval) { minind = i; // update the index of the list minval = arr[i][ptr[i]]; } // find maximum value among all the list elements pointing by the ptr[] array if (ptr[i] < n && arr[i][ptr[i]] > maxval) { maxval = arr[i][ptr[i]]; } } // if any list exhaust we will not get any better answer, so break the while loop if (flag == 1) { break ; } ptr[minind]++; // updating the minrange if ((maxval - minval) < minrange) { minel = minval; maxel = maxval; minrange = maxel - minel; } } document.write( "The smallest range is [" +minel+ ", " +maxel+ "]<br>" ); } // Driver program to test above function let arr = [ [4, 7, 9, 12, 15], [0, 8, 10, 14, 20], [6, 12, 16, 30, 50] ] let k = arr.length; findSmallestRange(arr, N, k); // This code is contributed by unknown2108 </script> |
Output:
The smallest range is [6 8]
- Complexity Analysis:
- Time complexity : O(n * k2), to find the maximum and minimum in an array of length k the time required is O(k), and to traverse all the k arrays of length n (in worst case), the time complexity is O(n*k), then the total time complexity is O(n*k2).
- Space complexity: O(k), an extra array is required of length k so the space complexity is O(k)
Efficient approach: The approach remains the same but the time complexity can be reduced by using min-heap or priority queue. Min heap can be used to find the maximum and minimum value in logarithmic time or log k time instead of linear time. Rest of the approach remains the same.
- Algorithm:
- create a Min heap to store k elements, one from each array and a variable minrange initialized to a maximum value and also keep a variable max to store the maximum integer.
- Initially put the first element of each element from each list and store the maximum value in max.
- Repeat the following steps until at least one list exhausts :
- To find the minimum value or min, use the top or root of the Min heap which is the minimum element.
- Now update the minrange if current (max-min) is less than minrange.
- remove the top or root element from priority queue and insert the next element from the list which contains the min element and update the max with the new element inserted.
- Implementation:
C++
// C++ program to finds out smallest range that includes // elements from each of the given sorted lists. #include <bits/stdc++.h> using namespace std; #define N 5 // A min heap node struct MinHeapNode { // The element to be stored int element; // index of the list from which the element is taken int i; // index of the next element to be picked from list int j; }; // Prototype of a utility function to swap two min heap nodes void swap(MinHeapNode* x, MinHeapNode* y); // A class for Min Heap class MinHeap { // pointer to array of elements in heap MinHeapNode* harr; // size of min heap int heap_size; public : // Constructor: creates a min heap of given size MinHeap(MinHeapNode a[], int size); // to heapify a subtree with root at given index void MinHeapify( int ); // to get index of left child of node at index i int left( int i) { return (2 * i + 1); } // to get index of right child of node at index i int right( int i) { return (2 * i + 2); } // to get the root MinHeapNode getMin() { return harr[0]; } // to replace root with new node x and heapify() new root void replaceMin(MinHeapNode x) { harr[0] = x; MinHeapify(0); } }; // Constructor: Builds a heap from a // given array a[] of given size MinHeap::MinHeap(MinHeapNode a[], int size) { heap_size = size; harr = a; // store address of array int i = (heap_size - 1) / 2; while (i >= 0) { MinHeapify(i); i--; } } // A recursive method to heapify a subtree with root at // given index. This method assumes that the subtrees // are already heapified void MinHeap::MinHeapify( int i) { int l = left(i); int r = right(i); int smallest = i; if (l < heap_size && harr[l].element < harr[i].element) smallest = l; if (r < heap_size && harr[r].element < harr[smallest].element) smallest = r; if (smallest != i) { swap(harr[i], harr[smallest]); MinHeapify(smallest); } } // This function takes an k sorted lists in the form of // 2D array as an argument. It finds out smallest range // that includes elements from each of the k lists. void findSmallestRange( int arr[][N], int k) { // Create a min heap with k heap nodes. Every heap node // has first element of an list int range = INT_MAX; int min = INT_MAX, max = INT_MIN; int start, end; MinHeapNode* harr = new MinHeapNode[k]; for ( int i = 0; i < k; i++) { // Store the first element harr[i].element = arr[i][0]; // index of list harr[i].i = i; // Index of next element to be stored // from list harr[i].j = 1; // store max element if (harr[i].element > max) max = harr[i].element; } // Create the heap MinHeap hp(harr, k); // Now one by one get the minimum element from min // heap and replace it with next element of its list while (1) { // Get the minimum element and store it in output MinHeapNode root = hp.getMin(); // update min min = hp.getMin().element; // update range if (range > max - min + 1) { range = max - min + 1; start = min; end = max; } // Find the next element that will replace current // root of heap. The next element belongs to same // list as the current root. if (root.j < N) { root.element = arr[root.i][root.j]; root.j += 1; // update max element if (root.element > max) max = root.element; } // break if we have reached end of any list else break ; // Replace root with next element of list hp.replaceMin(root); } cout << "The smallest range is " << "[" << start << " " << end << "]" << endl; ; } // Driver program to test above functions int main() { int arr[][N] = { { 4, 7, 9, 12, 15 }, { 0, 8, 10, 14, 20 }, { 6, 12, 16, 30, 50 } }; int k = sizeof (arr) / sizeof (arr[0]); findSmallestRange(arr, k); return 0; } |
Java
// Java program to find out smallest // range that includes elements from // each of the given sorted lists. class GFG { // A min heap node static class Node { // The element to be stored int ele; // index of the list from which // the element is taken int i; // index of the next element // to be picked from list int j; Node( int a, int b, int c) { this .ele = a; this .i = b; this .j = c; } } // A class for Min Heap static class MinHeap { Node[] harr; // array of elements in heap int size; // size of min heap // Constructor: creates a min heap // of given size MinHeap(Node[] arr, int size) { this .harr = arr; this .size = size; int i = (size - 1 ) / 2 ; while (i >= 0 ) { MinHeapify(i); i--; } } // to get index of left child // of node at index i int left( int i) { return 2 * i + 1 ; } // to get index of right child // of node at index i int right( int i) { return 2 * i + 2 ; } // to heapify a subtree with // root at given index void MinHeapify( int i) { int l = left(i); int r = right(i); int small = i; if (l < size && harr[l].ele < harr[i].ele) small = l; if (r < size && harr[r].ele < harr[small].ele) small = r; if (small != i) { swap(small, i); MinHeapify(small); } } void swap( int i, int j) { Node temp = harr[i]; harr[i] = harr[j]; harr[j] = temp; } // to get the root Node getMin() { return harr[ 0 ]; } // to replace root with new node x // and heapify() new root void replaceMin(Node x) { harr[ 0 ] = x; MinHeapify( 0 ); } } // This function takes an k sorted lists // in the form of 2D array as an argument. // It finds out smallest range that includes // elements from each of the k lists. static void findSmallestRange( int [][] arr, int k) { int range = Integer.MAX_VALUE; int min = Integer.MAX_VALUE; int max = Integer.MIN_VALUE; int start = - 1 , end = - 1 ; int n = arr[ 0 ].length; // Create a min heap with k heap nodes. // Every heap node has first element of an list Node[] arr1 = new Node[k]; for ( int i = 0 ; i < k; i++) { Node node = new Node(arr[i][ 0 ], i, 1 ); arr1[i] = node; // store max element max = Math.max(max, node.ele); } // Create the heap MinHeap mh = new MinHeap(arr1, k); // Now one by one get the minimum element // from min heap and replace it with // next element of its list while ( true ) { // Get the minimum element and // store it in output Node root = mh.getMin(); // update min min = root.ele; // update range if (range > max - min + 1 ) { range = max - min + 1 ; start = min; end = max; } // Find the next element that will // replace current root of heap. // The next element belongs to same // list as the current root. if (root.j < n) { root.ele = arr[root.i][root.j]; root.j++; // update max element if (root.ele > max) max = root.ele; } // break if we have reached // end of any list else break ; // Replace root with next element of list mh.replaceMin(root); } System.out.print( "The smallest range is [" + start + " " + end + "]" ); } // Driver Code public static void main(String[] args) { int arr[][] = { { 4 , 7 , 9 , 12 , 15 }, { 0 , 8 , 10 , 14 , 20 }, { 6 , 12 , 16 , 30 , 50 } }; int k = arr.length; findSmallestRange(arr, k); } } // This code is contributed by nobody_cares |
C#
// C# program to find out smallest // range that includes elements from // each of the given sorted lists. using System; using System.Collections.Generic; class GFG { // A min heap node public class Node { // The element to be stored public int ele; // index of the list from which // the element is taken public int i; // index of the next element // to be picked from list public int j; public Node( int a, int b, int c) { this .ele = a; this .i = b; this .j = c; } } // A class for Min Heap public class MinHeap { // array of elements in heap public Node[] harr; // size of min heap public int size; // Constructor: creates a min heap // of given size public MinHeap(Node[] arr, int size) { this .harr = arr; this .size = size; int i = (size - 1) / 2; while (i >= 0) { MinHeapify(i); i--; } } // to get index of left child // of node at index i int left( int i) { return 2 * i + 1; } // to get index of right child // of node at index i int right( int i) { return 2 * i + 2; } // to heapify a subtree with // root at given index void MinHeapify( int i) { int l = left(i); int r = right(i); int small = i; if (l < size && harr[l].ele < harr[i].ele) small = l; if (r < size && harr[r].ele < harr[small].ele) small = r; if (small != i) { swap(small, i); MinHeapify(small); } } void swap( int i, int j) { Node temp = harr[i]; harr[i] = harr[j]; harr[j] = temp; } // to get the root public Node getMin() { return harr[0]; } // to replace root with new node x // and heapify() new root public void replaceMin(Node x) { harr[0] = x; MinHeapify(0); } } // This function takes an k sorted lists // in the form of 2D array as an argument. // It finds out smallest range that includes // elements from each of the k lists. static void findSmallestRange( int [, ] arr, int k) { int range = int .MaxValue; int min = int .MaxValue; int max = int .MinValue; int start = -1, end = -1; int n = arr.GetLength(0); // Create a min heap with k heap nodes. // Every heap node has first element of an list Node[] arr1 = new Node[k]; for ( int i = 0; i < k; i++) { Node node = new Node(arr[i, 0], i, 1); arr1[i] = node; // store max element max = Math.Max(max, node.ele); } // Create the heap MinHeap mh = new MinHeap(arr1, k); // Now one by one get the minimum element // from min heap and replace it with // next element of its list while ( true ) { // Get the minimum element and // store it in output Node root = mh.getMin(); // update min min = root.ele; // update range if (range > max - min + 1) { range = max - min + 1; start = min; end = max; } // Find the next element that will // replace current root of heap. // The next element belongs to same // list as the current root. if (root.j < n) { root.ele = arr[root.i, root.j]; root.j++; // update max element if (root.ele > max) max = root.ele; } else break ; // break if we have reached // end of any list // Replace root with next element of list mh.replaceMin(root); } Console.Write( "The smallest range is [" + start + " " + end + "]" ); } // Driver Code public static void Main(String[] args) { int [, ] arr = { { 4, 7, 9, 12, 15 }, { 0, 8, 10, 14, 20 }, { 6, 12, 16, 30, 50 } }; int k = arr.GetLength(0); findSmallestRange(arr, k); } } // This code is contributed by Rajput-Ji |
Javascript
<script> // Javascript program to find out smallest // range that includes elements from // each of the given sorted lists. class Node { constructor(a, b, c) { this .ele = a; this .i = b; this .j = c; } } // A class for Min Heap class MinHeap { // Array of elements in heap harr; // Size of min heap size; // Constructor: creates a min heap // of given size constructor(arr,size) { this .harr = arr; this .size = size; let i = Math.floor((size - 1) / 2); while (i >= 0) { this .MinHeapify(i); i--; } } // To get index of left child // of node at index i left(i) { return 2 * i + 1; } // To get index of right child // of node at index i right(i) { return 2 * i + 2; } // To heapify a subtree with // root at given index MinHeapify(i) { let l = this .left(i); let r = this .right(i); let small = i; if (l < this .size && this .harr[l].ele < this .harr[i].ele) small = l; if (r < this .size && this .harr[r].ele < this .harr[small].ele) small = r; if (small != i) { this .swap(small, i); this .MinHeapify(small); } } swap(i, j) { let temp = this .harr[i]; this .harr[i] = this .harr[j]; this .harr[j] = temp; } // To get the root getMin() { return this .harr[0]; } // To replace root with new node x // and heapify() new root replaceMin(x) { this .harr[0] = x; this .MinHeapify(0); } } // This function takes an k sorted lists // in the form of 2D array as an argument. // It finds out smallest range that includes // elements from each of the k lists. function findSmallestRange(arr, k) { let range = Number.MAX_VALUE; let min = Number.MAX_VALUE; let max = Number.MIN_VALUE; let start = -1, end = -1; let n = arr[0].length; // Create a min heap with k heap nodes. // Every heap node has first element of an list let arr1 = new Array(k); for (let i = 0; i < k; i++) { let node = new Node(arr[i][0], i, 1); arr1[i] = node; // Store max element max = Math.max(max, node.ele); } // Create the heap let mh = new MinHeap(arr1, k); // Now one by one get the minimum element // from min heap and replace it with // next element of its list while ( true ) { // Get the minimum element and // store it in output let root = mh.getMin(); // Update min min = root.ele; // Update range if (range > max - min + 1) { range = max - min + 1; start = min; end = max; } // Find the next element that will // replace current root of heap. // The next element belongs to same // list as the current root. if (root.j < n) { root.ele = arr[root.i][root.j]; root.j++; // Update max element if (root.ele > max) max = root.ele; } // Break if we have reached // end of any list else break ; // Replace root with next element of list mh.replaceMin(root); } document.write( "The smallest range is [" + start + " " + end + "]" ); } // Driver Code let arr = [ [ 4, 7, 9, 12, 15 ], [ 0, 8, 10, 14, 20 ], [ 6, 12, 16, 30, 50 ] ]; let k = arr.length; findSmallestRange(arr, k); // This code is contributed by rag2127 </script> |
Output:
The smallest range is [6 8]
- Complexity Analysis:
- Time complexity : O(n * k *log k).
To find the maximum and minimum in a Min Heap of length k the time required is O(log k), and to traverse all the k arrays of length n (in the worst case), the time complexity is O(n*k), then the total time complexity is O(n * k *log k). - Space complexity: O(k).
The priority queue will store k elements so the space complexity of O(k)
- Time complexity : O(n * k *log k).
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