SOUJITH JADI
@SOUJITH_JADI
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Joined July 2025
Day 34:.Problem: Allocate Minimum Number of Pages .Approach: Used Binary Search on the answer. Checked feasibility with a helper function to see if pages can be allocated within the limit of students. Balanced efficiency with correctness by narrowing search range. @geeksforgeeks
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Day 33:.Problem: K-th element of two sorted arrays š¢.Approach: Used a two-pointer merge strategy, similar to merge sort. Traversed both arrays while counting elements until reaching the K-th element. Optimized for early termination once K is found. ā”.@geeksforgeeks š
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Day 32:.Problem:Search in Rotated Sorted Array .Approach:Applied modified binary search to handle the rotation. At each step, determined which half is sorted and checked if the target lies there.Reduced search space logarithmically until the target was found or confirmed absent
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Day 31:.Problem:Peak Element š.Approach:Used binary search to find any index where the element is greater than both its neighbors.Checked boundary cases by treating out-of-bound neighbors as negative infinity. Efficiently narrowed down to the peak in O(log n). @geeksforgeeks
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DAY 30 WITH #gfg160 .Problem: Find Minimum in Rotated Sorted Array š.Approach: Binary search twist! Compared middle with the rightmost element to decide which half to explore. This shrinks the search space in O(log n) and finds the min efficiently. š.@geeksforgeeks š
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IT'SS Day 29:.Problem:Kth Missing Positive Number in a Sorted Array.Approach:Stored array elements in a HashSet for O(1) lookups, then iterated from 1 upward, counting missing numbers until reaching the kth. Simple, direct, and efficient enough for constraints! š.@geeksforgeeks
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Day 28:.Problem: Aggressive Cows š.Approach: Applied Binary Search on the answer ā checked if cows can be placed at least mid distance apart using a greedy check. Adjusted low/high to maximize the minimum distance. Efficient O(N log M) solution! ā”š.@geeksforgeeks š
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Day 27:.Problem: Number of Occurrence.Approach: Used Binary Search twice ā once to find the first occurrence and once for the last occurrence of the target in a sorted array. The count = (last - first + 1). Efficient O(log N) lookup! ā”š.@geeksforgeeks š
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Day 26:.Problem: Count Inversions.Approach: Applied Modified Merge Sort to count how many times elements in the right half are smaller than the left half while merging. Efficient O(N log N) solution instead of brute force O(NĀ²)! š§ š”.@geeksforgeeks š
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It's Day 25: #gfg160 .Problem: Merge Without Extra Space.Approach: Used Gap Method (Shell Sort Variant) ā gradually reduce the comparison distance (gap) and swap out-of-place elements across both arrays. Efficient in-place merging with no extra space! š„.@geeksforgeeks šŖ
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