Leetcode 74. Search a 2D Matrix

 Problem Link

Write an efficient algorithm that searches for a value target in an m x n integer matrix matrix. This matrix has the following properties:

• Integers in each row are sorted from left to right.
• The first integer of each row is greater than the last integer of the previous row.

In this problem we can find the appropriate row number by using binary search

Then similarly we can find the appropriate row number by using binary search.

So, this solution requires two binary search.

Code:

class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        int m=matrix.size();
        int n=matrix[0].size();
        int loRow=0, hiRow= m-1;
        while(hiRow>=loRow){
            int midRow = (hiRow+loRow)/2;
            if(matrix[midRow][0]>target){
                hiRow = midRow-1;
            }
            else{
                loRow = midRow+1;
            }
        }
        // cout << hiRow << " "<< loRow << endl;
        int hiCol=n-1;
        int loCol=0;
        if(hiRow<0)return false;
        while(hiCol>=loCol){
            int midCol = (hiCol+loCol)/2;
            if(matrix[hiRow][midCol]>target){
                hiCol = midCol-1;
            }
            else{
                loCol = midCol+1;
            }
        }
        if(hiCol<0)return false;
        if(matrix[hiRow][hiCol]==target) return true;
        else return false;
    }
}; 

Count inversions in an array || Merge Sort

Problem link

Problem Statement: Given an array of N integers, count the inversion of the array.

What is an inversion of an array? Definition: for all i & j < size of array, if i < j then you have to find pair (A[i],A[j]) such that A[j] < A[i].

for better explanation click 

>>Basically this problem required a direct implementation of merge sort. 

>>The number of inversion is the number of time it finds the array mismatched during a merge sort process. In a merge sort process we divide the array into two halves. until we have the unit of the array elements. Then just comparing the unit variables and put them in right places and merge back to the array.

Code

#include <bits/stdc++.h> 
long long Merge(long long *arr, long long *temp, long long left, long long mid, long long right){
    long long cnt = 0;
    long long i = left;
    long long j = mid;
    long long k = left;
    while((i<=mid-1) && (j<=right)){
        if(arr[i]<=arr[j]){
            temp[k++] = arr[i++];
        }
        else{
            temp[k++] = arr[j++];
            cnt = cnt + (mid-i);
        }
    }
    
    // leftovers
    while(i<=mid-1){
        temp[k++] = arr[i++];
    }
    while(j<=right){
        temp[k++] = arr[j++];
    }
    
    for( i = left; i<=right;i++){
        arr[i] = temp[i];
    }
    return cnt;
}

long long mergeSort(long long *arr, long long *temp, long long left,long long right){
    long long mid, cnt = 0;
    if(right>left){
        mid = (left+right)/2;
        cnt += mergeSort(arr,temp,left,mid);
        cnt += mergeSort(arr,temp,mid+1,right);
        cnt += Merge(arr,temp,left,mid+1,right);
    }
    return cnt;
}
long long getInversions(long long *arr, int n){
    long long temp[n];
    return mergeSort(arr,temp,0,n-1);
}

DP || Leetcode 5. Longest Palindromic Substring

Problem Link

In this problem, there will be given a string, we have to find the longest palindromic substring of this string.

For this purpose, we have to check all the substring we can generate. From them the longest one which is palindrome will be the answer. In the naive approach, surely will cross the time limit.

DP approach: Suppose we have a string "cababa". Lets simulate this string by building a table of 6*6.

The cell value (i,j) = 0, means starting from index i to index j of the string doesn't have a palindrome

The cell value (i,j) = 1, means starting from index i to index j of the string have a palindrome, and its size is (j-i+1)

1. A single character is a palindrome itself. so, we will fill all the diagonal(i,i) as 1. 

So, the table looks like Fig :1

                                           Fig[1]
2. We will check all of 2 size substring, they formed palindrome[if the string characters are same] or not.

    for (0,1)>> characters are 'c' and 'a' so its not a palindrome we will put 0 at (0,1) 

    for (1,2)>> characters are 'a' and 'b' so its not a palindrome we will put 0 at (1,2)

    for (2,3)>> characters are 'b' and 'a' so its not a palindrome we will put 0 at (2,3)

    ....

    ....

 

3.Then we will check if the start index i and the end index j is the same character. if they are same then we will check between them all the characters are palindrome or not. In other word mathematically we will say a range palindrome if and only if,
    s[i]==s[j] and table[i+1][j-1]==1

table[i+1][j-1] represents middle substring, if you dont catch it, you can understand this by doing some paper works.

Finally, our looks like Fig 2.

                                            Fig[2]

Code:

class Solution {
public:
    int dp[1000][1000];
    string longestPalindrome(string s) {
        int ans=1;
        int start=0,end=0;
        for(int i=0;i<(int)s.size();i++){
            dp[i][i] =1; 
        }
        for(int i=0;i<(int)s.size()-1;i++){
            if(s[i]==s[i+1]){ dp[i][i+1] = 1;ans = 2;start=i;end=i+1;}
            else dp[i][i+1] = 0;
        }
        
        for(int j=2;j<(int)s.size();j++){
            for(int i=0;i<(int)s.size()-2;i++){
                if(s[i]==s[j] && dp[i+1][j-1]==1){
                    dp[i][j] = 1;
                    ans = max(ans,j-i+1);
                    if(ans==j-i+1){
                        start = i;
                        end = j;
                    }
                }
            }
        }
        string ansString = "";
        for(int i=start;i<=end;i++){
            ansString.push_back(s[i]);
        }
        return ansString;
    }
};