javacpp

 Graph                         (4) -------------(3)                    /                                   \                 /                                         \           (5)------------(1)-----------(2) in This Above graph 4 connected to 3 and 5. Then 3 connected to 2 and 4 and so on. Graph are Used To Various Algorithms and Technology for example, Shortest Path Algorithm.               1.Graph.               2.bidirectional Graph.                3.Weighted Graph. Bidirectional  Graph               We Can Travel Both Side . Like Two Way Road. S======>E   and E<=====S (start to end and end to start) vice versa. Weighted Graph               Weighted Graph Have Length Of Path. Like Km Is Roadways. Properties vertex edge Vertex is a defined as start or end point. Edge is a connection between start and end points     import java.util.*; class graph< T >{ HashMap < T , LinkedList < T >> g= new HashMap < T , LinkedList < T >>(); publ

What is  Symmetric  Matrix? A=[1 2]       [2 1] is a Square Matrix . When We Modify The Given Matrix Row Into Column And Column Into Row. if Modified and Original Matrix Array Are Same It Known As Symmetric Matrix. If A=A'     A=[1 2]                              A'=[1 2]            [2 1]   is Equal To       [2  1] Other Wise Given Matrix is Non Symmetric. The Follow Simple Python Code For Find The Given Array Is Symmetric or Not. arr=[[1,2],[2,1]] n=len(arr[0]) m=len(arr) isTrue=1 for i in range(0,n):     for j in range(0,m):         if(arr[i][j]!=arr[j][i]):             isTrue=0             break; print(isTrue)  output: 1 Time Complexity o(n^2). Comment Your solutions.

public static void main( String [] args){ int [] arr={ 1 , 2 , 3 , 4 , 5 , 6 }; int n=arr.length; int sum= 0 ; for ( int i= 0 ;i<n;i++){ sum+=arr[i]; } System.out.println(sum); } } Java Programming /* Sum Of Given Array Elements Time Complexity O(n) */   class sumOfArray{

In this article we are discuss about Basic Problem in Programming. How To Reverse a Array Elements. Reverse a given array elements. For example1, [1,2,3,4,5] Output [5,4,3,2,1] example2, Input [3,5,1,2,4] Output [4,2,1,5,3] 1) With Extra Space (Copy Arrar)  We can simple make a same size of array and copy from given array elements to new array variable. Its takes o(n) Time Complexity. And more Space To Run Programme. class arrayReverse{ public static void main( String [] args){ int [] arr={ 1 , 2 , 3 , 4 , 5 }; int n=arr.length; int [] brr= new int [n]; for ( int i= 0 ;i<n;i++){ brr[i]=arr[n-i- 1 ]; } for ( int j= 0 ;j<n;j++){ System.out.println(brr[j]); } } } 2)With Out Extra Space. We can Swap The Array Element One By One. Like First Element And Last. Second Element And Second Last Until Reach Middle Element. class arrayReverseWithoutExtraSpace{ public static void main( Strin

  Find and remove Array Elements Which Appear More Then Once.(Remove Duplicate Elements From Sorted Array). For Example1, Input, [4,4,3,3,1] Output, [4,3,1] For Example2, Input, [5,5,4,3,1] Output, [5,4,3,1] Solution 1)Iterate the array elements. 2)Intalize The Index Counter(j). 3)Check If The Current And Next Element Are Same(i not equals i+1). 4)If Its Not Equal Store arr[i] into jth index. 5)do Until Itrate Whole Array Elements. How to find number of Array Elements Without Dublicate Array Elements. J+1 Elements Non Dublicate Array  Elements. Java class arrayDublicateWithNoExtraSpace{ public static void main( String [] args){ int [] arr={ 1 , 2 , 2 , 3 , 4 , 5 , 5 , 6 , 6 , 6 , 7 }; int n=arr.length; int j= 0 ; for ( int i= 0 ;i<n- 1 ;i++){ if (arr[i]!=arr[i+ 1 ]){ arr[j]=arr[i]; j+= 1 ; } } //Buffer Last Element Of Given Array if (arr[n- 1 ]!=arr[n

  class arrayDublicate{ //Remove Array Dublicate With Extra Space public static void main( String [] args){ int [] arr={ 1 , 2 , 2 , 3 , 4 , 5 , 5 , 6 }; int n=arr.length; int [] copyArr= new int [n]; int j= 0 ; for ( int i= 0 ;i<n- 1 ;i++){ if (arr[i]!=arr[i+ 1 ]){ copyArr[j]=arr[i]; j+= 1 ; } } for ( int i= 0 ;i<j;i++){ System.out.println(copyArr[i]); } } }

class longestCommonSubstring{ public static int lcs( String str, String str2, int n, int m){ if (m< 0 ||n< 0 ){ return 0 ; } else if (str.charAt(n)==str2.charAt(m)){ return 1 +lcs(str,str2,n- 1 ,m- 1 ); } else { return max(lcs(str,str2,n- 1 ,m),lcs(str,str2,n,m- 1 )); } } public static int max( int a, int b){ if (a>b){ return a; } else { return b; } } public static void main( String [] args){ String str1= "abcdefgh" ; String str2= "eghabcd" ; System.out.println(lcs(str1,str2,str1.length()- 1 ,str2.length()- 1 )); } }   output: 4