.js

//动态规划，不在乎过程，只在乎结果，后面依赖前面结果
//最后一步：到结尾
//子问题：目标是ryr形式，翻牌3次，dp[0]翻r,dp[1]翻y,dp[2]翻r
//状态转移方程：
// dp[0][i] = dp[0][i-1] + incRed
// dp[1][i] = Math.min(dp[0][i - 1], dp[1][i - 1]) + incYellow
// dp[2][i] = Math.min(dp[1][i - 1], dp[2][i - 1]) + incRed
//初始条件：
// dp[0][0] = leaves[0] === 'r' ? 0 : 1
// dp[1][0] = Number.MAX_VALUE
// dp[2][0] = Number.MAX_VALUE
// dp[2][1] = Number.MAX_VALUE
//计算顺序：0 到 length

var minimumOperations = function (leaves) {
  let dp = []
  dp[0] = []
  dp[1] = []
  dp[2] = []

  dp[0][0] = leaves[0]==='r' ? 0 : 1
  dp[1][0] = Number.MAX_VALUE
  dp[2][0] = Number.MAX_VALUE
  dp[2][1] = Number.MAX_VALUE
  let l = leaves.length
  for (let i = 1; i < l; i++) {
    let incRed = leaves[i] === 'y' ? 1: 0
    let incYellow = leaves[i] === 'y' ? 0:1
    dp[0][i] = dp[0][i-1] + incRed
    dp[1][i] = Math.min(dp[0][i - 1], dp[1][i - 1]) + incYellow
    if(i>=2){
      dp[2][i] = Math.min(dp[1][i - 1], dp[2][i - 1]) + incRed
    }
  }
  return dp[2][l-1]
};

console.log(minimumOperations("rrryyyrryyyrr"))