The intended solution uses Dynamic Programming.

Let <code>dp[i]</code> denote the minimum number of coins, such that we bought <code>i^th</code> fruit and acquired all the fruits in the range <code>[i...n]</code>.

<code>dp[i] = min(dp[i], dp[j] + prices[i]) </code>, where <code>j</code> is in the range <code>[i + 1, i + 1 + i]</code>.