LeetCode 1536 | Minimum Swaps to Arrange a Binary Grid | Greedy + Simulation | Java | Medium

LeetCode 1536 – Minimum Swaps to Arrange a Binary Grid (Java)

Today's LeetCode Daily

You are given an n x n binary grid.

In one step:
You can swap two adjacent rows.

Goal:
Make the grid valid.

A grid is valid if:
All cells above the main diagonal are 0.

Core Insight:

For row i,
we need at least n minus 1 minus i
trailing zeros at the end of that row.

Why?
Because positions above the diagonal
must contain zeros.

Approach Used:

Step 1:
For each row,
count trailing zeros.

Step 2:
For row i,
required trailing zeros =
n minus 1 minus i.

Step 3:
Find the first row j
below i that satisfies requirement.

Step 4:
If no such row exists,
return -1.

Step 5:
Bubble that row upward
by swapping adjacent rows.
Count swaps.

Key Idea:
Greedy row selection
plus adjacent swap simulation.

Time Complexity: O(n squared)
Space Complexity: O(n)

Problem Link:
https://leetcode.com/problems/minimum-swaps-to-arrange-a-binary-grid/?envType=daily-question&envId=2026-03-02

Java Solution (GitHub):
https://github.com/Amandf/LeetCode-Solutions-/blob/main/1536-minimum-swaps-to-arrange-a-binary-grid/1536-minimum-swaps-to-arrange-a-binary-grid.java

More daily Java solutions.
Clean code. No mic. Pure logic.

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