# No BFS

• Sample array {4,1,1,1,3,1,1}
The first jump is 4. This implies that starting from 0 until 4th index, the min number of jumps to be taken is 1. While we travel from index 0 to 4, we also track the next max jump.
At index 0, max is 4
At index 1, max is 3 (max(3,1))
At index 2, max is 2 (max(2,1))
At index 3, max is 1 (max(1,1))
At index 4, max is 3 (max(0,3))
Total number of min steps to reach 0 to 4th index is 1.
At index 5, max is 2 (max(2,1))
At index 6, max is 1 (max(1,1))
Total number of min steps to reach 0 to 6th index is 2.

``````public int jump(int[] jumps) {
int j = jumps[0];
int max = jumps[0];
int steps = 0;
int i = 0;
// loop through the array
while(i < jumps.length-1) {
// loop until the current steps is not 0. While looping, track the next max steps.
while(j>0 && i < jumps.length-1) {
j--;
max = Math.max(jumps[++i],max-1);
}
steps++;
j = max;
if(j==0) break;
}
return (i==jumps.length-1)?steps:0;
}
``````

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