Given a matrix where every row is sorted in increasing order. Write a function that finds and returns a common element in all rows. If there is no common element, then returns -1.
Example:
Input: mat[4][5] = { {1, 2, 3, 4, 5},
{2, 4, 5, 8, 10},
{3, 5, 7, 9, 11},
{1, 3, 5, 7, 9},
};
Output: 5
A O(m*n*n) simple solution is to take every element of first row and search it in all other rows, till we find a common element. Time complexity of this solution is O(m*n*n) where m is number of rows and n is number of columns in given matrix. This can be improved to O(m*n*Logn) if we use Binary Search instead of linear search.
We can solve this problem in O(mn) time using the approach similar to merge of Merge Sort. The idea is to start from the last column of every row. If elements at all last columns are same, then we found the common element. Otherwise we find the minimum of all last columns. Once we find a minimum element, we know that all other elements in last columns cannot be a common element, so we reduce last column index for all rows except for the row which has minimum value. We keep repeating these steps till either all elements at current last column don't become same, or a last column index reaches 0.
Below is the implementation of above idea.
C++ // A C++ program to find a common element in all rows of a // row wise sorted array #include <bits/stdc++.h> using namespace std; // Specify number of rows and columns #define M 4 #define N 5 // Returns common element in all rows of mat[M][N]. If there is no // common element, then -1 is returned int findCommon(int mat[M][N]) { // An array to store indexes of current last column int column[M]; int min_row; // To store index of row whose current // last element is minimum // Initialize current last element of all rows int i; for (i = 0; i < M; i++) column[i] = N - 1; min_row = 0; // Initialize min_row as first row // Keep finding min_row in current last column, till either // all elements of last column become same or we hit first column. while (column[min_row] >= 0) { // Find minimum in current last column for (i = 0; i < M; i++) { if (mat[i][column[i]] < mat[min_row][column[min_row]]) min_row = i; } // eq_count is count of elements equal to minimum in current last // column. int eq_count = 0; // Traverse current last column elements again to update it for (i = 0; i < M; i++) { // Decrease last column index of a row whose value is more // than minimum. if (mat[i][column[i]] > mat[min_row][column[min_row]]) { if (column[i] == 0) return -1; column[i] -= 1; // Reduce last column index by 1 } else eq_count++; } // If equal count becomes M, return the value if (eq_count == M) return mat[min_row][column[min_row]]; } return -1; } // Driver Code int main() { int mat[M][N] = { { 1, 2, 3, 4, 5 }, { 2, 4, 5, 8, 10 }, { 3, 5, 7, 9, 11 }, { 1, 3, 5, 7, 9 }, }; int result = findCommon(mat); if (result == -1) cout << "No common element"; else cout << "Common element is " << result; return 0; } // This code is contributed // by Akanksha Rai // A C program to find a common element in all rows of a // row wise sorted array #include <stdio.h> // Specify number of rows and columns #define M 4 #define N 5 // Returns common element in all rows of mat[M][N]. If there is no // common element, then -1 is returned int findCommon(int mat[M][N]) { // An array to store indexes of current last column int column[M]; int min_row; // To store index of row whose current // last element is minimum // Initialize current last element of all rows int i; for (i = 0; i < M; i++) column[i] = N - 1; min_row = 0; // Initialize min_row as first row // Keep finding min_row in current last column, till either // all elements of last column become same or we hit first column. while (column[min_row] >= 0) { // Find minimum in current last column for (i = 0; i < M; i++) { if (mat[i][column[i]] < mat[min_row][column[min_row]]) min_row = i; } // eq_count is count of elements equal to minimum in current last // column. int eq_count = 0; // Traverse current last column elements again to update it for (i = 0; i < M; i++) { // Decrease last column index of a row whose value is more // than minimum. if (mat[i][column[i]] > mat[min_row][column[min_row]]) { if (column[i] == 0) return -1; column[i] -= 1; // Reduce last column index by 1 } else eq_count++; } // If equal count becomes M, return the value if (eq_count == M) return mat[min_row][column[min_row]]; } return -1; } // driver program to test above function int main() { int mat[M][N] = { { 1, 2, 3, 4, 5 }, { 2, 4, 5, 8, 10 }, { 3, 5, 7, 9, 11 }, { 1, 3, 5, 7, 9 }, }; int result = findCommon(mat); if (result == -1) printf("No common element"); else printf("Common element is %d", result); return 0; } // A Java program to find a common // element in all rows of a // row wise sorted array class GFG { // Specify number of rows and columns static final int M = 4; static final int N = 5; // Returns common element in all rows // of mat[M][N]. If there is no // common element, then -1 is // returned static int findCommon(int mat[][]) { // An array to store indexes // of current last column int column[] = new int[M]; // To store index of row whose current // last element is minimum int min_row; // Initialize current last element of all rows int i; for (i = 0; i < M; i++) column[i] = N - 1; // Initialize min_row as first row min_row = 0; // Keep finding min_row in current last column, till either // all elements of last column become same or we hit first column. while (column[min_row] >= 0) { // Find minimum in current last column for (i = 0; i < M; i++) { if (mat[i][column[i]] < mat[min_row][column[min_row]]) min_row = i; } // eq_count is count of elements equal to minimum in current last // column. int eq_count = 0; // Traverse current last column elements again to update it for (i = 0; i < M; i++) { // Decrease last column index of a row whose value is more // than minimum. if (mat[i][column[i]] > mat[min_row][column[min_row]]) { if (column[i] == 0) return -1; // Reduce last column index by 1 column[i] -= 1; } else eq_count++; } // If equal count becomes M, // return the value if (eq_count == M) return mat[min_row][column[min_row]]; } return -1; } // Driver code public static void main(String[] args) { int mat[][] = { { 1, 2, 3, 4, 5 }, { 2, 4, 5, 8, 10 }, { 3, 5, 7, 9, 11 }, { 1, 3, 5, 7, 9 } }; int result = findCommon(mat); if (result == -1) System.out.print("No common element"); else System.out.print("Common element is " + result); } } // This code is contributed by Anant Agarwal. # Python 3 program to find a common element # in all rows of a row wise sorted array # Specify number of rows # and columns M = 4 N = 5 # Returns common element in all rows # of mat[M][N]. If there is no common # element, then -1 is returned def findCommon(mat): # An array to store indexes of # current last column column = [N - 1] * M min_row = 0 # Initialize min_row as first row # Keep finding min_row in current last # column, till either all elements of # last column become same or we hit first column. while (column[min_row] >= 0): # Find minimum in current last column for i in range(M): if (mat[i][column[i]] < mat[min_row][column[min_row]]): min_row = i # eq_count is count of elements equal # to minimum in current last column. eq_count = 0 # Traverse current last column elements # again to update it for i in range(M): # Decrease last column index of a row # whose value is more than minimum. if (mat[i][column[i]] > mat[min_row][column[min_row]]): if (column[i] == 0): return -1 column[i] -= 1 # Reduce last column # index by 1 else: eq_count += 1 # If equal count becomes M, return the value if (eq_count == M): return mat[min_row][column[min_row]] return -1 # Driver Code if __name__ == "__main__": mat = [[1, 2, 3, 4, 5], [2, 4, 5, 8, 10], [3, 5, 7, 9, 11], [1, 3, 5, 7, 9]] result = findCommon(mat) if (result == -1): print("No common element") else: print("Common element is", result) # This code is contributed by ita_c // A C# program to find a common // element in all rows of a // row wise sorted array using System; class GFG { // Specify number of rows and columns static int M = 4; static int N = 5; // Returns common element in all rows // of mat[M][N]. If there is no // common element, then -1 is // returned static int findCommon(int[, ] mat) { // An array to store indexes // of current last column int[] column = new int[M]; // To store index of row whose // current last element is minimum int min_row; // Initialize current last element // of all rows int i; for (i = 0; i < M; i++) column[i] = N - 1; // Initialize min_row as first row min_row = 0; // Keep finding min_row in current // last column, till either all // elements of last column become // same or we hit first column. while (column[min_row] >= 0) { // Find minimum in current // last column for (i = 0; i < M; i++) { if (mat[i, column[i]] < mat[min_row, column[min_row]]) min_row = i; } // eq_count is count of elements // equal to minimum in current // last column. int eq_count = 0; // Traverse current last column // elements again to update it for (i = 0; i < M; i++) { // Decrease last column index // of a row whose value is more // than minimum. if (mat[i, column[i]] > mat[min_row, column[min_row]]) { if (column[i] == 0) return -1; // Reduce last column index // by 1 column[i] -= 1; } else eq_count++; } // If equal count becomes M, // return the value if (eq_count == M) return mat[min_row, column[min_row]]; } return -1; } // Driver code public static void Main() { int[, ] mat = { { 1, 2, 3, 4, 5 }, { 2, 4, 5, 8, 10 }, { 3, 5, 7, 9, 11 }, { 1, 3, 5, 7, 9 } }; int result = findCommon(mat); if (result == -1) Console.Write("No common element"); else Console.Write("Common element is " + result); } } // This code is contributed by Sam007. <script> // A Javascript program to find a common // element in all rows of a // row wise sorted array // Specify number of rows and columns let M = 4; let N = 5; // Returns common element in all rows // of mat[M][N]. If there is no // common element, then -1 is // returned function findCommon(mat) { // An array to store indexes // of current last column let column=new Array(M); // To store index of row whose current // last element is minimum let min_row; // Initialize current last element of all rows let i; for (i = 0; i < M; i++) column[i] = N - 1; // Initialize min_row as first row min_row = 0; // Keep finding min_row in current // last column, till either // all elements of last column become // same or we hit first column. while (column[min_row] >= 0) { // Find minimum in current last column for (i = 0; i < M; i++) { if (mat[i][column[i]] < mat[min_row][column[min_row]]) min_row = i; } // eq_count is count of elements equal to // minimum in current last // column. let eq_count = 0; // Traverse current last column // elements again to update it for (i = 0; i < M; i++) { // Decrease last column index of a row whose value is more // than minimum. if (mat[i][column[i]] > mat[min_row][column[min_row]]) { if (column[i] == 0) return -1; // Reduce last column index by 1 column[i] -= 1; } else eq_count++; } // If equal count becomes M, // return the value if (eq_count == M) return mat[min_row][column[min_row]]; } return -1; } // Driver Code let mat = [[1, 2, 3, 4, 5], [2, 4, 5, 8, 10], [3, 5, 7, 9, 11], [1, 3, 5, 7, 9]]; let result = findCommon(mat) if (result == -1) { document.write("No common element"); } else { document.write("Common element is ", result); } // This code is contributed by rag2127 </script> OutputCommon element is 5
Time complexity: O(M x N).
Auxiliary Space: O(M)
Explanation for working of above code
Let us understand working of above code for following example.
Initially entries in last column array are N-1, i.e., {4, 4, 4, 4}
{1, 2, 3, 4, 5},
{2, 4, 5, 8, 10},
{3, 5, 7, 9, 11},
{1, 3, 5, 7, 9},
The value of min_row is 0, so values of last column index for rows with value greater than 5 is reduced by one. So column[] becomes {4, 3, 3, 3}.
{1, 2, 3, 4, 5},
{2, 4, 5, 8, 10},
{3, 5, 7, 9, 11},
{1, 3, 5, 7, 9},
The value of min_row remains 0 and value of last column index for rows with value greater than 5 is reduced by one. So column[] becomes {4, 2, 2, 2}.
{1, 2, 3, 4, 5},
{2, 4, 5, 8, 10},
{3, 5, 7, 9, 11},
{1, 3, 5, 7, 9},
The value of min_row remains 0 and value of last column index for rows with value greater than 5 is reduced by one. So column[] becomes {4, 2, 1, 2}.
{1, 2, 3, 4, 5},
{2, 4, 5, 8, 10},
{3, 5, 7, 9, 11},
{1, 3, 5, 7, 9},
Now all values in current last columns of all rows is same, so 5 is returned.
A Hashing Based Solution
We can also use hashing. This solution works even if the rows are not sorted. It can be used to print all common elements.
Step1: Create a Hash Table with all key as distinct elements
of row1. Value for all these will be 0.
Step2:
For i = 1 to M-1
For j = 0 to N-1
If (mat[i][j] is already present in Hash Table)
If (And this is not a repetition in current row.
This can be checked by comparing HashTable value with
row number)
Update the value of this key in HashTable with current
row number
Step3: Iterate over HashTable and print all those keys for
which value = M
C++ // C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Specify number of rows and columns #define M 4 #define N 5 // Returns common element in all rows of mat[M][N]. If there is no // common element, then -1 is returned int findCommon(int grid[M][N]) { // A hash map to store count of elements unordered_map<int, int> cnt; int i, j; for (i = 0; i < M; i++) { // Increment the count of first // element of the row cnt[grid[i][0]]++; // Starting from the second element // of the current row for (j = 1; j < N; j++) { // If current element is different from // the previous element i.e. it is appearing // for the first time in the current row if (grid[i][j] != grid[i][j - 1]) cnt[grid[i][j]]++; } } // Find element having count equal to number of rows for (auto ele : cnt) { if (ele.second == M) return ele.first; } // No such element found return -1; } // Driver Code int main() { int mat[M][N] = { { 1, 2, 3, 4, 5 }, { 2, 4, 5, 8, 10 }, { 3, 5, 7, 9, 11 }, { 1, 3, 5, 7, 9 }, }; int result = findCommon(mat); if (result == -1) cout << "No common element"; else cout << "Common element is " << result; return 0; } // Java implementation of the approach import java.util.*; class GFG { // Specify number of rows and columns static int M = 4; static int N = 5; // Returns common element in all rows of mat[M][N]. // If there is no common element, then -1 is returned static int findCommon(int mat[][]) { // A hash map to store count of elements HashMap<Integer, Integer> cnt = new HashMap<Integer, Integer>(); int i, j; for (i = 0; i < M; i++) { // Increment the count of first // element of the row if(cnt.containsKey(mat[i][0])) { cnt.put(mat[i][0], cnt.get(mat[i][0]) + 1); } else { cnt.put(mat[i][0], 1); } // Starting from the second element // of the current row for (j = 1; j < N; j++) { // If current element is different from // the previous element i.e. it is appearing // for the first time in the current row if (mat[i][j] != mat[i][j - 1]) if(cnt.containsKey(mat[i][j])) { cnt.put(mat[i][j], cnt.get(mat[i][j]) + 1); } else { cnt.put(mat[i][j], 1); } } } // Find element having count // equal to number of rows for (Map.Entry<Integer, Integer> ele : cnt.entrySet()) { if (ele.getValue() == M) return ele.getKey(); } // No such element found return -1; } // Driver Code public static void main(String[] args) { int mat[][] = {{ 1, 2, 3, 4, 5 }, { 2, 4, 5, 8, 10 }, { 3, 5, 7, 9, 11 }, { 1, 3, 5, 7, 9 }}; int result = findCommon(mat); if (result == -1) System.out.println("No common element"); else System.out.println("Common element is " + result); } } // This code is contributed by Rajput-Ji # Python3 implementation of the approach from collections import defaultdict # Specify number of rows and columns M = 4 N = 5 # Returns common element in all rows of # mat[M][N]. If there is no # common element, then -1 is returned def findCommon(grid): global M global N # A hash map to store count of elements cnt = dict() cnt = defaultdict(lambda: 0, cnt) i = 0 j = 0 while (i < M ): # Increment the count of first # element of the row cnt[grid[i][0]] = cnt[grid[i][0]] + 1 j = 1 # Starting from the second element # of the current row while (j < N ) : # If current element is different from # the previous element i.e. it is appearing # for the first time in the current row if (grid[i][j] != grid[i][j - 1]): cnt[grid[i][j]] = cnt[grid[i][j]] + 1 j = j + 1 i = i + 1 # Find element having count equal to number of rows for ele in cnt: if (cnt[ele] == M): return ele # No such element found return -1 # Driver Code mat = [[1, 2, 3, 4, 5 ], [2, 4, 5, 8, 10], [3, 5, 7, 9, 11], [1, 3, 5, 7, 9 ],] result = findCommon(mat) if (result == -1): print("No common element") else: print("Common element is ", result) # This code is contributed by Arnab Kundu // C# implementation of the approach using System; using System.Collections.Generic; class GFG { // Specify number of rows and columns static int M = 4; static int N = 5; // Returns common element in all rows of mat[M,N]. // If there is no common element, then -1 is returned static int findCommon(int [,]grid) { // A hash map to store count of elements Dictionary<int, int> cnt = new Dictionary<int, int>(); int i, j; for (i = 0; i < M; i++) { // Increment the count of first // element of the row if(cnt.ContainsKey(grid[i, 0])) { cnt[grid[i, 0]]= cnt[grid[i, 0]] + 1; } else { cnt.Add(grid[i, 0], 1); } // Starting from the second element // of the current row for (j = 1; j < N; j++) { // If current element is different from // the previous element i.e. it is appearing // for the first time in the current row if (grid[i, j] != grid[i, j - 1]) if(cnt.ContainsKey(mat[i, j])) { cnt[grid[i, j]]= cnt[grid[i, j]] + 1; } else { cnt.Add(grid[i, j], 1); } } } // Find element having count // equal to number of rows foreach(KeyValuePair<int, int> ele in cnt) { if (ele.Value == M) return ele.Key; } // No such element found return -1; } // Driver Code public static void Main(String[] args) { int [,]mat = {{ 1, 2, 3, 4, 5 }, { 2, 4, 5, 8, 10 }, { 3, 5, 7, 9, 11 }, { 1, 3, 5, 7, 9 }}; int result = findCommon(mat); if (result == -1) Console.WriteLine("No common element"); else Console.WriteLine("Common element is " + result); } } // This code is contributed by 29AjayKumar <script> // Javascript implementation of the approach // Specify number of rows and columns let M = 4; let N = 5; // Returns common element in all rows of mat[M][N]. // If there is no common element, then -1 is returned function findCommon(mat) { // A hash map to store count of elements let cnt = new Map(); let i, j; for (i = 0; i < M; i++) { // Increment the count of first // element of the row if(cnt.has(mat[i][0])) { cnt.set(mat[i][0],cnt.get(mat[i][0])+1); } else { cnt.set(mat[i][0],1); } // Starting from the second element // of the current row for (j = 1; j < N; j++) { // If current element is different from // the previous element i.e. it is appearing // for the first time in the current row if (mat[i][j] != mat[i][j - 1]) { if(cnt.has(mat[i][j])) { cnt.set(mat[i][j], cnt.get(mat[i][j]) + 1); } else { cnt.set(mat[i][j], 1); } } } } // Find element having count // equal to number of rows for( let [key, value] of cnt.entries()) { if(value == M) return key; } // No such element found return -1; } // Driver Code let mat = [[1, 2, 3, 4, 5 ], [2, 4, 5, 8, 10], [3, 5, 7, 9, 11], [1, 3, 5, 7, 9 ],] let result = findCommon(mat); if (result == -1) document.write("No common element"); else document.write("Common element is " + result); // This code is contributed by avanitrachhadiya2155 </script> OutputCommon element is 5
Time complexity: O(n*m) under the assumption that search and insert in HashTable take O(1) time
Auxiliary Space: O(n) due to unordered_map.
Thanks to Nishant for suggesting this solution in a comment below.
Exercise: Given n sorted arrays of size m each, find all common elements in all arrays in O(mn) time.
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