Return true if you can reach the last index starting from index 0 with maximum jump lengths given by nums[i].

Example 1:

• Input: nums = [2,3,1,1,4]
• Output: true

Constraints:

• 1 <= nums.length <= 10^4

Greedy backward goal post shift.

Greedy Reachability

graph TD
    Problem["Problem of Size N"] --> Sub["Break into Subproblems DP[i]"]
    Sub --> Base["Base Cases: DP[0], DP[1]"]
    Base --> Trans["State Transition: DP[i] = f(DP[i-1], DP[i-2], ...)"]
    Trans --> Table["Fill DP Table / Variables"]
    Table --> Result["Return DP[N]"]

function canJump(nums) {
  let goal = nums.length - 1;
  for (let i = nums.length - 1; i >= 0; i--) {
    if (i + nums[i] >= goal) goal = i;
  }
  return goal === 0;
}

• Time Complexity: O(n)
• Space Complexity: O(1)
• Explanation: Backward greedy goal post update.

function canJump(nums) {
  let goal = nums.length - 1;
  for (let i = nums.length - 1; i >= 0; i--) {
    if (i + nums[i] >= goal) goal = i;
  }
  return goal === 0;
}

• Time Complexity: O(n)
• Space Complexity: O(1)
• Explanation: Greedy single pass.

1. Initialize State: Setup necessary pointers, dynamic programming arrays, or hash maps.
2. Iterate & Evaluate: Process the input according to the boundary conditions.
3. Update & Return: Compute the optimal answer and return early or at termination.

Iterate backwards starting from end; if current index plus jump distance reaches goal, shift goal to current index.

1. Track target goal from right to left.

👉 Solve this problem interactively in the DSA Lab