Description
Given an array of integers arr, you are initially positioned at the first index of the array.
In one step you can jump from index i to index:
i + 1where:i + 1 < arr.length.i - 1where:i - 1 >= 0.jwhere:arr[i] == arr[j]andi != j.
Return the minimum number of steps to reach the last index of the array.
Notice that you can not jump outside of the array at any time.
Example 1:
Input: arr = [100,-23,-23,404,100,23,23,23,3,404] Output: 3 Explanation: You need three jumps from index 0 --> 4 --> 3 --> 9. Note that index 9 is the last index of the array.
Example 2:
Input: arr = [7] Output: 0 Explanation: Start index is the last index. You do not need to jump.
Example 3:
Input: arr = [7,6,9,6,9,6,9,7] Output: 1 Explanation: You can jump directly from index 0 to index 7 which is last index of the array.
Constraints:
1 <= arr.length <= 5 * 104-108 <= arr[i] <= 108
Solutions
This code solves a minimum jumps problem using BFS (Breadth-First Search). It first builds a map that stores all indices for each unique value in the array, then explores positions level-by-level starting from index 0. At each position, you can jump to three types of locations: any other index with the same value (found via the map), the left neighbor (pos - 1), or the right neighbor (pos + 1). The algorithm tracks visited positions using a seen array to avoid cycles and deletes each value from the map after exploring it to avoid redundant checks. It counts steps until reaching the final position (n - 1), returning the number of jumps needed, or -1 if unreachable.
/**
* @param {number[]} arr
* @return {number}
*/
var minJumps = function(arr) {
const n = arr.length;
const map = new Map();
for (let i = 0; i < n; i++) {
if (!map.has(arr[i])) {
map.set(arr[i], []);
}
map.get(arr[i]).push(i);
}
let stack = [0];
const seen = Array(n).fill(false);
seen[0] = true;
let steps = 0;
while (stack.length > 0) {
const next = [];
for (const pos of stack) {
if (pos === n - 1) {
return steps;
}
const mapArr = map.get(arr[pos]);
if (mapArr) {
for (const con of mapArr) {
if (con !== pos && !seen[con]) {
seen[con] = true;
next.push(con);
}
}
map.delete(arr[pos]);
}
const back = pos - 1;
if (back >= 0 && !seen[back]) {
seen[back] = true;
next.push(back);
}
const forw = pos + 1;
if (forw < n && !seen[forw]) {
seen[forw] = true;
next.push(forw);
}
}
stack = next;
steps++;
}
return -1;
};