Given a matrix containing lower alphabetical characters only, we need to count number of palindromic paths in given matrix. A path is defined as a sequence of cells starting from top-left cell and ending at bottom-right cell. We are allowed to move to right and down only from current cell.
Examples:
Input : mat[][] = {"aaab”,
"baaa”
“abba”}
Output : 3
Number of palindromic paths are 3 from top-left to
bottom-right.
aaaaaa (0, 0) -> (0, 1) -> (1, 1) -> (1, 2) ->
(1, 3) -> (2, 3)
aaaaaa (0, 0) -> (0, 1) -> (0, 2) -> (1, 2) ->
(1, 3) -> (2, 3)
abaaba (0, 0) -> (1, 0) -> (1, 1) -> (1, 2) ->
(2, 2) -> (2, 3)
We can solve this problem recursively, we start from two corners of a palindromic path(top-left and bottom right). In each recursive call, we maintain a state which will constitute two cells one from starting and one from end which should be equal for palindrome property. If at a state, both cell characters are equal then we call recursively with all possible movements in both directions.
As this can lead to solving same subproblem multiple times, we have taken a map memo in below code which stores the calculated result with key as indices of starting and ending cell so if subproblem with same starting and ending cell is called again, result will be returned by memo directly instead of recalculating again.
Implementation:
CPP
// C++ program to get number of palindrome// paths in matrix#include <bits/stdc++.h>using namespace std;#define R 3#define C 4// struct to represent state of recursion// and key of mapstruct cells{ // indices of front cell int rs, cs; // indices of end cell int re, ce; cells(int rs, int cs, int re, int ce) : rs(rs) , cs(cs) , re(re) , ce(ce) { } // operator overloading to compare two // cells which rs needed for map bool operator<(const cells& other) const { return ((rs != other.rs) || (cs != other.cs) || (re != other.re) || (ce != other.ce)); }};// recursive method to return number// of palindromic paths in matrix// (rs, cs) ==> Indices of current cell // from a starting point (First Row)// (re, ce) ==> Indices of current cell // from a ending point (Last Row)// memo ==> To store results of // already computed problemsint getPalindromicPathsRecur(char mat[R][C], int rs, int cs, int re, int ce, map<cells, int>& memo){ // Base Case 1 : if any index rs out of boundary, // return 0 if (rs < 0 || rs >= R || cs < 0 || cs >= C) return 0; if (re < 0 || re < rs || ce < 0 || ce < cs) return 0; // Base case 2 : if values are not equal // then palindrome property rs not satisfied, // so return 0 if (mat[rs][cs] != mat[re][ce]) return 0; // If we reach here, then matrix cells are same. // Base Case 3 : if indices are adjacent then // return 1 if (abs((rs - re) + (cs - ce)) <= 1) return 1; // if result rs precalculated, return from map if (memo.find(cells(rs, cs, re, ce)) != memo.end()) return memo[cells(rs, cs, re, ce)]; int ret = 0; // Initialize result // calling recursively for all possible movements ret += getPalindromicPathsRecur(mat, rs + 1, cs, re - 1, ce, memo); ret += getPalindromicPathsRecur(mat, rs + 1, cs, re, ce - 1, memo); ret += getPalindromicPathsRecur(mat, rs, cs + 1, re - 1, ce, memo); ret += getPalindromicPathsRecur(mat, rs, cs + 1, re, ce - 1, memo); // storing the calculated result in map memo[cells(rs, cs, re, ce)] = ret; return ret;}// method returns number of palindromic paths in matrixint getPalindromicPaths(char mat[R][C]){ map<cells, int> memo; return getPalindromicPathsRecur(mat, 0, 0, R - 1, C - 1, memo);}// Driver code int main(){ char mat[R][C] = { 'a', 'a', 'a', 'b', 'b', 'a', 'a', 'a', 'a', 'b', 'b', 'a' }; printf("%d", getPalindromicPaths(mat)); return 0;} |
Java
// Java program to get number of palindrome paths in matriximport java.util.HashMap;// Cells class definitionclass Cells { int rs; int cs; int re; int ce; // Constructor public Cells(int rs, int cs, int re, int ce) { this.rs = rs; this.cs = cs; this.re = re; this.ce = ce; } // Equals method to compare two Cells objects @Override public boolean equals(Object o) { if (o instanceof Cells) { Cells other = (Cells)o; return (rs == other.rs) && (cs == other.cs) && (re == other.re) && (ce == other.ce); } return false; } // hashCode() function to calculate hash @Override public int hashCode() { return (rs * 100 + cs) * 100 + (re * 100 + ce); }}class PalindromicPaths { // Rows and Columns in the matrix static int R = 3; static int C = 4; // method to return number of palindromic paths in // matrix static int getPalindromicPaths(char[][] mat) { HashMap<Cells, Integer> memo = new HashMap<>(); R = mat.length; C = mat[0].length; return getPalindromicPathsRecur(mat, 0, 0, R - 1, C - 1, memo); } // recursive method to return number of palindromic // paths in matrix static int getPalindromicPathsRecur(char[][] mat, int rs, int cs, int re, int ce, HashMap<Cells, Integer> memo) { // Base Case 1 : if any index is out of boundary, // return 0 if (rs < 0 || rs >= R || cs < 0 || cs >= C) { return 0; } if (re < 0 || re < rs || ce < 0 || ce < cs) { return 0; } // Base case 2 : if values are not equal then // palindrome property is not satisfied, so return 0 if (mat[rs][cs] != mat[re][ce]) { return 0; } // If we reach here, then matrix cells are same. // Base Case 3 : if indices are adjacent then return // 1 if (Math.abs((rs - re) + (cs - ce)) <= 1) { return 1; } Cells key = new Cells(rs, cs, re, ce); // if result is precalculated, return from map if (memo.containsKey(key)) { return memo.get(key); } int ret = 0; // Initialize result // calling recursively for all possible movements ret += getPalindromicPathsRecur(mat, rs + 1, cs, re - 1, ce, memo); ret += getPalindromicPathsRecur(mat, rs + 1, cs, re, ce - 1, memo); ret += getPalindromicPathsRecur(mat, rs, cs + 1, re - 1, ce, memo); ret += getPalindromicPathsRecur( mat, rs, cs + 1, re - 1, ce - 1, memo); // storing the calculated result in map memo.put(key, ret); return ret; } // Driver code public static void main(String[] args) { char[][] mat = { { 'a', 'a', 'a' }, { 'b', 'b', 'a' }, { 'a', 'a', 'a' }, { 'b', 'b', 'a' } }; System.out.println(getPalindromicPaths(mat)); }}// This code is contributed by phasing17. |
Python3
# Python3 program to get number of palindrome# paths in matrixR = 3C = 4# struct to represent state of recursion# and key of mapclass Cells: def __init__(self, rs, cs, re, ce): self.rs = rs self.cs = cs self.re = re self.ce = ce def __eq__(self, other): if other: return (self.rs, self.cs, self.re, self.ce) == (other.rs, other.cs, other.re, other.ce) return False def __hash__(self): return hash((self.rs, self.cs, self.re, self.ce))# recursive method to return number# of palindromic paths in matrix# (rs, cs) ==> Indices of current cell # from a starting point (First Row)# (re, ce) ==> Indices of current cell # from a ending point (Last Row)# memo ==> To store results of # already computed problemsdef getPalindromicPathsRecur(mat, rs, cs, re, ce, memo): # Base Case 1 : if any index rs out of boundary, # return 0 if rs < 0 or rs >= R or cs < 0 or cs >= C: return 0 if re < 0 or re < rs or ce < 0 or ce < cs: return 0 # Base case 2 : if values are not equal # then palindrome property rs not satisfied, # so return 0 if mat[rs][cs] != mat[re][ce]: return 0 # If we reach here, then matrix cells are same. # Base Case 3 : if indices are adjacent then # return 1 if abs((rs - re) + (cs - ce)) <= 1: return 1 key = Cells(rs, cs, re, ce) # if result rs precalculated, return from map if key in memo: return memo[key] ret = 0 # Initialize result # calling recursively for all possible movements ret += getPalindromicPathsRecur(mat, rs + 1, cs, re - 1, ce, memo) ret += getPalindromicPathsRecur(mat, rs + 1, cs, re, ce - 1, memo) ret += getPalindromicPathsRecur(mat, rs, cs + 1, re - 1, ce, memo) ret += getPalindromicPathsRecur(mat, rs, cs + 1, re - 1, ce - 1, memo) # storing the calculated result in map memo[key] = ret return ret# method returns number of palindromic paths in matrixdef getPalindromicPaths(mat): memo = {} R = len(mat) C = len(mat[0]) res = getPalindromicPathsRecur(mat, 0, 0, R - 1, C - 1, memo) return res# Driver code if __name__ == "__main__": mat = [['a', 'a', 'a'], ['b', 'b', 'a'], ['a', 'a', 'a'], ['b', 'b', 'a']] print(getPalindromicPaths(mat)) # This code is contributed by phasing17. |
C#
// C# program to get number of palindrome paths in matrixusing System;using System.Collections.Generic;// Cells class definitionclass Cells { int rs; int cs; int re; int ce; // Constructor public Cells(int rs, int cs, int re, int ce) { this.rs = rs; this.cs = cs; this.re = re; this.ce = ce; } // Equals method to compare two Cells objects public override bool Equals(Object o) { if (o is Cells) { Cells other = (Cells)o; return (rs == other.rs) && (cs == other.cs) && (re == other.re) && (ce == other.ce); } return false; } // hashCode() function to calculate hash public override int GetHashCode() { return (rs * 100 + cs) * 100 + (re * 100 + ce); }}class PalindromicPaths { // Rows and Columns in the matrix static int R = 3; static int C = 4; // method to return number of palindromic paths in // matrix static int GetPalindromicPaths(char[][] mat) { Dictionary<Cells, int> memo = new Dictionary<Cells, int>(); R = mat.Length; C = mat[0].Length; return GetPalindromicPathsRecur(mat, 0, 0, R - 1, C - 1, memo); } // recursive method to return number of palindromic // paths in matrix static int GetPalindromicPathsRecur(char[][] mat, int rs, int cs, int re, int ce, Dictionary<Cells, int> memo) { // Base Case 1 : if any index is out of boundary, // return 0 if (rs < 0 || rs >= R || cs < 0 || cs >= C) { return 0; } if (re < 0 || re < rs || ce < 0 || ce < cs) { return 0; } // Base case 2 : if values are not equal then // palindrome property is not satisfied, so return 0 if (mat[rs][cs] != mat[re][ce]) { return 0; } // If we reach here, then matrix cells are same. // Base Case 3 : if indices are adjacent then return // 1 if (Math.Abs((rs - re) + (cs - ce)) <= 1) { return 1; } Cells key = new Cells(rs, cs, re, ce); // if result is precalculated, return from map if (memo.ContainsKey(key)) { return memo[key]; } int ret = 0; // Initialize result // calling recursively for all possible movements ret += GetPalindromicPathsRecur(mat, rs + 1, cs, re - 1, ce, memo); ret += GetPalindromicPathsRecur(mat, rs + 1, cs, re, ce - 1, memo); ret += GetPalindromicPathsRecur(mat, rs, cs + 1, re - 1, ce, memo); ret += GetPalindromicPathsRecur( mat, rs, cs + 1, re - 1, ce - 1, memo); // storing the calculated result in map memo[key] = ret; return ret; } // Driver code public static void Main(string[] args) { char[][] mat = { new char[] { 'a', 'a', 'a' }, new char[] { 'b', 'b', 'a' }, new char[] { 'a', 'a', 'a' }, new char[] { 'b', 'b', 'a' } }; Console.WriteLine(GetPalindromicPaths(mat)); }}// This code is contributed by phasing17. |
Javascript
// JavaScript code to get number of palindrome paths in matrixconst R = 3;const C = 4;// class to represent state of recursion// and key of mapclass Cells { constructor(rs, cs, re, ce) { this.rs = rs; this.cs = cs; this.re = re; this.ce = ce; } // custom equality check equals(other) { if (other) { return ( this.rs === other.rs && this.cs === other.cs && this.re === other.re && this.ce === other.ce ); } return false; }}// map to store results of already computed problemsconst memo = new Map();// recursive method to return number of palindromic paths in matrix// (rs, cs) ==> Indices of current cell from a starting point (First Row)// (re, ce) ==> Indices of current cell from a ending point (Last Row)function getPalindromicPathsRecur(mat, rs, cs, re, ce) { // Base Case 1 : if any index rs out of boundary, return 0 if (rs < 0 || rs >= R || cs < 0 || cs >= C) { return 0; } if (re < 0 || re < rs || ce < 0 || ce < cs) { return 0; } // Base case 2 : if values are not equal then palindrome property rs not satisfied, so return 0 if (mat[rs][cs] !== mat[re][ce]) { return 0; } // If we reach here, then matrix cells are same. // Base Case 3 : if indices are adjacent then return 1 if (Math.abs((rs - re) + (cs - ce)) <= 1) { return 1; } const key = new Cells(rs, cs, re, ce); // if result rs precalculated, return from map if (memo.has(key)) { return memo.get(key); } let ret = 0; // Initialize result // calling recursively for all possible movements ret += getPalindromicPathsRecur(mat, rs + 1, cs, re - 1, ce); ret += getPalindromicPathsRecur(mat, rs + 1, cs, re, ce - 1); ret += getPalindromicPathsRecur(mat, rs, cs + 1, re - 1, ce); ret += getPalindromicPathsRecur(mat, rs, cs + 1, re - 1, ce - 1); // storing the calculated result in map memo.set(key, ret); return ret;}// method returns number of palindromic paths in matrixfunction getPalindromicPaths(mat) { const R = mat.length; const C = mat[0].length; const res = getPalindromicPathsRecur(mat, 0, 0, R - 1, C - 1); return res;}// Driver codelet mat = [['a', 'a', 'a'], ['b', 'b', 'a'], ['a', 'a', 'a'], ['b', 'b', 'a']]console.log(getPalindromicPaths(mat))// This code is contributed by phasing17. |
Output
3
Time Complexity : O((R x C)2)
Space Complexity : O((R x C)2) due to use of memoization map.
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