Increasing Number Triangle Starting from 0 in JavaScript

What You’ll Learn

The increasing number triangle using i + j prints 0, 1 2, 2 3 4, … — a natural follow-up after Program 33’s i + j - 1 triangle starting from 1. This tutorial covers the i + j formula, nested loops, a live preview, worked JavaScript examples, edge cases, and complexity.

📝 Problem & Approach

Given max_n = 5, print a zero-based left-shifted increasing triangle: for each row i, print j = 0..i values of i + j separated by spaces.

// max_n = 5 (i runs 0..5)
const max_n = 5;
for (let i = 0; i <= max_n; i++) {
  let line = "";
  for (let j = 0; j <= i; j++) {
    line += (i + j) + " ";
  }
  console.log(line);
}

Inputs & Outputs

Minimal workflow

for i from 0 to max_n:
    for j from 0 to i:
        print (i + j) + space
    print newline

Approach comparison

🔮 Live Preview

Choose a max_n value between 2 and 9 and draw the increasing triangle in the browser.

Press "Draw pattern".

Examples Gallery

Three complete JavaScript programs — fixed max_n, user input, and a smaller trace demo. Click View Output to reveal sample console results, or Try it Yourself to run the code live.

const max_n = 5;

for (let i = 0; i <= max_n; i++) {
  let line = "";
  for (let j = 0; j <= i; j++) {
    line += (i + j) + " ";
  }
  console.log(line);
}

0
1 2
2 3 4
3 4 5 6
4 5 6 7 8
5 6 7 8 9 10

const maxInput = prompt("Enter max i:");
const max_n = parseInt(maxInput, 10);

if (!Number.isFinite(max_n) || max_n < 0) {
  console.log("Please enter a non-negative integer.");
} else {
  for (let i = 0; i <= max_n; i++) {
    let line = "";
    for (let j = 0; j <= i; j++) {
      line += (i + j) + " ";
    }
    console.log(line);
  }
}

Enter max i: 2
0
1 2
2 3 4

const max_n = 2;

for (let i = 0; i <= max_n; i++) {
  let line = "";
  for (let j = 0; j <= i; j++) {
    line += (i + j) + " ";
  }
  console.log(line);
}

0
1 2
2 3 4

🔎 Worked Walkthrough — max_n = 5

Trace each outer-loop value of i, inner-loop range, values printed, and full row output.

Prints per row = i + 1 — total prints = (max_n+1)(max_n+2)/2 when i runs 0..max_n.

🎯 Practice Problems

Try these variations to lock in the pattern.

Notes

⏱️ Time and Space Complexity

💡 Best Practices