
![he most important DP pattern you'll ever learn. 0/1 Knapsack — every interview has a version of it. The recurrence is just two choices: Skip the item → dp[i][w] = dp[i-1][w] Include the item → dp[i][w](https://image.pictame.com/img/669898888_18044994104587857_3921722089787642932_n.jpg?hash=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)

![Brute force checks every pair → O(n²). HashMap does it in O(n). The trick: Build freq map in one pass For each key, check if key + k exists in the map If yes → pairs += freq[key] × freq[key+k] That's](https://image.pictame.com/img/660978872_18042882596587857_1615295072721945166_n.jpg?hash=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)








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🧠 Breaking down code, visually. ⚡ DSA • Java • LeetCode • Dev 🎯 1 concept clearer every day.