Cheapest Flights Within K Stops LeetCode solution in JavaScript

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/**
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 * @param {number} n
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 * @param {number[][]} flights
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 * @param {number} src
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 * @param {number} dst
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 * @param {number} k
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 * @return {number}
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 */
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const MAX_PRICE = Math.pow(10, 8);
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var findCheapestPrice = function (n, flights, src, dst, k) {
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  let prev = [];
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  let step = 0;
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  let curr = [];
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  for (let i = 0; i < n; i++) {
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    prev[i] = MAX_PRICE;
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  }
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  prev[src] = 0;
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  while (step - 1 < k) {
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    curr = [...prev];
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    let isAnyChange = false;
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    for (let i = 0; i < flights.length; i++) {
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      let [src, dst, p] = flights[i];
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      if (prev[src] >= MAX_PRICE) continue;
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      let totalCostToReachDst = prev[src] + p;
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      curr[dst] = Math.min(curr[dst], totalCostToReachDst);
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    }
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    step++;
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    prev = curr;
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  }
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  return prev[dst] >= MAX_PRICE ? -1 : prev[dst];
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};