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HTML

              
                <h1>Find first value greater than given number
</h1>
<div class="alert alert-note">
  All problems solved in JavaScript given in this <a href="https://www.youtube.com/watch?v=GU7DpgHINWQ&list=PLl0KD3g-oDOHpWRyyGBUJ9jmul0lUOD80">binary search YouTube video.</a>

  <p>
  <h3>Rule of thumb on Binary Search</h3>
  <p>
    Divide your array in set of TRUE and FALSE values then find the First occurrence of TRUE or the last occurrence of TRUE. Binary Search Array should be partially false and partially true. We need to find the boundary between them.
  </p>
  </p>

  <p>
  <h3>
    Lower Bound Problem
  </h3>
  Find the first value Greater than or Equal to 4.
  <img src="https://i.imgur.com/MOQYomw.png" alt="">

  <li>If the middle value is True then search in left side to find more true.</li>
  <li>If the middle value is False then search in right side to find True.</li>

  </p>

</div>
              
            
!

CSS

              
                .alert {
  padding: 20px;
  border-radius: 25px;
  margin-bottom: 15px;
}
img {
  width: 100%;
  height: auto;
}
.alert-tip {
  background-color: #def6dd;
}

.alert-note {
  background-color: #efd9fd;
  overflow-x: auto;
}
pre {
  overflow-x: auto;
  white-space: pre-wrap;
  white-space: -moz-pre-wrap;
  white-space: -pre-wrap;
  white-space: -o-pre-wrap;
  word-wrap: break-word;
}

              
            
!

JS

              
                /*
Lower Bound of 4 Values which are less than 4.
*/
function findFirstValue(array, data) {
  let low = 0;
  let high = array.length - 1;
  let mid;
  let result = -1;
  while (low <= high) {
    mid = low + Math.floor((high - low) / 2);
    if (array[mid] >= data) {
      result = array[mid];
      high = mid - 1; // since I need first occurrence so keep searching in the left sorted half of the array.
    } else low = mid + 1;
  }

  return result;
}
describe("Find First Value >= x in sorted array", () => {
  it("should work correctly #1", () => {
    expect(findFirstValue([2, 3, 5, 6, 8, 10, 12], 4)).toBe(5);
  });
});

              
            
!
999px

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