36 - Word Break - Dynamic Programming approach

@Rishi Srivastava Pseudo code: boolean[] dp = new boolean[() 1]; dp[()] = true; for (int i = () - 1; i is greater than or equal to 0; i--) { for (final String word : wordDict) { if (((i ()) is less than or equal to ()) && ((word, i))) { dp[i] = dp[i ()]; } if (dp[i]) { break; } } } return dp[0]; Github: Leetcode: