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Showing posts with the label LeetCode

Java Code to Partition an Array and return the maximized Sum of Pairs

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  Given an integer array  nums  of  2n  integers, group these integers into  n  pairs  (a 1 , b 1 ), (a 2 , b 2 ), ..., (a n , b n )  such that the sum of  min(a i , b i )  for all  i  is  maximized . Return  the maximized sum This is a leetcode problem , which we will try to solve today.  On analyzing the problem we can understand that we need to split or partition the array into pairs of 2, and then take the minimum value from a pair, such that the sum of these minimum values is maximum.  Ideally, the value can be maximum only if we are taking bigger numbers, but we have to take minimum values from a pair of Integers. So, to work around this problem, we will sort the array first. This way, when we make pairs, we can ensure that the minimum value of a pair can come out to be a bigger number. We will then take the sum and return the same.  /**   * Group an array of 2n integers , into n...

Java code to Split a String into Maximum number of Balanced Strings

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  We are given a string which contains only characters 'R' and 'L'. A balanced string is one which has equal number of 'R' and 'L' characters. The problem statement is to find the maximum number of balanced strings which can be extracted from an input String. Here is a link to the leetcode problem, and we'll try to solve it today. /**   * Split the String into maximum number of Balanced Strings   *   * @author computengine.com   * @param s   * @return  */ public int balancedStringSplit(String s) { int countR = 0 , countL = 0 , count = 0 ; // loop through the input string for ( int i = 0 ; i < s . length ( ) ; i + + ) { // increment if 'R' found if ( s . charAt ( i ) = = 'R' ) countR + + ; // increment if 'L' found if ( s . charAt ( i ) = = 'L' ) countL + + ; // increment counter when Number of 'R' match Number of 'L', and reset...

Java code to find Count of Pairs in Integer Array with Absolute Difference K

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Let's solve a LeetCode problem using Java.  Count Number of Pairs With Absolute Difference K You can go through the problem and try to understand what needs to be done. Try to write your own code first. We have solved the problem and its here for your reference. /**  * Count Number of Pairs With Absolute Difference K  * @author computengine.com  * @param nums  * @param k  * @return  */ public int countKDifference ( int [ ] nums , int k ) { int count = 0 ; for ( int i = 0 ; i < nums . length ; i + + ) { for ( int j = i + 1 ; j < nums . length ; j + + ) { if ( Math . abs ( nums [ i ] - nums [ j ] ) = = k ) { count + + ; } } } return count ; } If you look closely, we have looped through the Integer Array, and compared two elements of the arrays and find the difference between them. Now the difference can be positive or negative but it has to be equal to the input K. So we'v...