Tomcat Devs

Java · DSA · Two Pointers ◐ Push every 0 to the end of an array while keeping the other values in their original order , in place, with no extra array. Watch the two pointers i and k rearrange it step by step. 01 Problem 02 Code 03 How it works 04 Tracer 05 Approaches 06 Revision 07 Quiz 01 The problem Given an array of N integers, move all the zeros to the end. The non-zero values must keep their relative order , and you should change the array in place . For [0, 1, 0, 3, 12] the answer is [1, 3, 12, 0, 0] . The order 1 → 3 → 12 is unchanged. Type an array to see what the program prints. Array (spaces or commas) Run 02 The code Two loops and one extra variable. k is the next position where a non-zero value should go. The first loop packs the non-zeros to the front, and the second loop fills the rest with zeros. Copy package Day...

Java · DSA · Arrays ◐ Find the second largest distinct element in an array in a single pass , with no sorting and no extra array. Watch largest and secondLargest update step by step on your own input. 01 Problem 02 Code 03 How it works 04 Tracer 05 Approaches 06 Revision 07 Quiz 01 The problem Given an array of N integers, print the second largest distinct value. Duplicates of the largest don't count. If there is no second distinct value, print -1 . For [10, 20, 20, 15, 5] , the largest is 20. The next distinct value below it is 15 . For [7, 7, 7] there is no second value, so the answer is -1 . Type an array to see what the program prints. Negative numbers work too. Array (spaces or commas) Run 02 The code Keep two variables while walking the array once: largest and secondLargest . A boolean, isSLargestFound , records whether a r...

Java · DSA · Hashing ◐ Add the lowest and highest frequency in an array. Solve it two ways, brute force and hashing , then watch both run step by step on your own input. 01 Problem 02 Brute force 03 Hashing 04 Tracer 05 Comparison 06 Revision 07 Quiz 01 The problem Given an array of N non-negative integers, find how often each value occurs. Print the minimum frequency + maximum frequency . For [1, 2, 2, 3, 3, 3, 3] , the frequencies are 1 → 1, 2 → 2 and 3 → 4. The lowest is 1 and the highest is 4, so the answer is 1 + 4 = 5 . Type an array to run both programs and see every frequency. Array (spaces or commas) Run 02 Approach 1: Brute force Walk the array with i . The first time a value appears, mark it in visited and count it with an inner j loop. Then update minFreq and maxFreq with that count. Copy import java.util. Sca...