nums array to handle edge cases.dp where dp[left][right] represents the maximum coins obtainable by bursting all balloons between left and right.n.left.right based on the current length and starting point.i between left and right, calculate the coins obtained by assuming the ith balloon is the last to burst.i plus the coins from bursting the ith balloon itself, considering the neighbors.dp[left][right] with the maximum coins obtained for all positions i.dp[1][n-2] as the final answer, excluding the fake balloons added in step 1.